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I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Figure 1 shows an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$.

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example, see Figure 2:

Nor does further restricting L to the right half and R to the left half. For example, see Figure 3:

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example, see Figure 4:

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction. See Figure 5:

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$, as shown in Figure 6:

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$, as shown in Figure 7:

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Figure 8 shows a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Figure 9 shows a heat map of the fractional assignments of cell to direction:

Figure 10 shows a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

Figure 11 shows a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions:

I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Figure 1 shows an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$.

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example, see Figure 2:

Nor does further restricting L to the right half and R to the left half. For example, see Figure 3:

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example, see Figure 4:

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction. See Figure 5:

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$, as shown in Figure 6:

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$, as shown in Figure 7:

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Figure 8 shows a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Figure 9 shows a heat map of the fractional assignments of cell to direction:

Figure 10 shows a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

Figure 11 shows a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions:

Figure 12 shows a solution for $n=50$ where each cell has the minimum number $34$ of arrows pointing to it:

I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Here's an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$.

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example:

Nor does further restricting L to the right half and R to the left half. For example:

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example:

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction:

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$:

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$:

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Here's a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Here's a heat map of the fractional assignments of cell to direction:

Here's a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

And here's a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions:

I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Figure 1 shows an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$. 

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example, see Figure 2: 

Nor does further restricting L to the right half and R to the left half. For example, see Figure 3: 

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example, see Figure 4: 

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction. See Figure 5: 

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$, as shown in Figure 6: 

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$, as shown in Figure 7: 

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Figure 8 shows a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Figure 9 shows a heat map of the fractional assignments of cell to direction:

Figure 10 shows a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

Figure 11 shows a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions:

I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Here's an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$.

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example:

Nor does further restricting L to the right half and R to the left half. For example:

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example:

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction:

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$:

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$:

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Here's a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Here's a heat map of the fractional assignments of cell to direction:

Here's a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

And here's a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions:

I used integer linear programming (in SAS) to compute the exact values for $n\in\{2,\dots,50\}$, and they match https://oeis.org/A248183. Let $C_{i,j,d} \subseteq [n] \times [n]$ be the set of cells that can point to cell $(i,j)$ via direction $d$. Let binary decision variable $x_{i,j,d}$ indicate whether the arrow in cell $(i,j)$ points in direction $d$, and let $z$ represent the smallest number of arrows pointing to a cell. The problem is to maximize $z$ subject to \begin{align} \sum_d x_{i,j,d} &= 1 &&\text{for all $(i,j)$} \tag1\\ z &\le \sum_{(u,v) \in C_{i,j,d}} x_{u,v,d} && \text{for all $(i,j)$} \tag2 \end{align} Constraint $(1)$ chooses exactly one arrow per cell. Constraint $(2)$ enforces the maximin objective.

Here's an optimal solution for $n=50$, with $r(50)=34$. Maybe it will suggest a pattern for general $n$.

Restricting U to the bottom half and D to the top half does not change the optimal values for $n\le 50$. For example:

Nor does further restricting L to the right half and R to the left half. For example:

For even $n \le 50$, the optimal values still do not change if you further restrict each direction to appear $n^2/4$ times. For example:

Forcing horizontal and vertical symmetry, and forcing triangle with side length $10$ ($>10$ loses optimality), as suggested by @Wolfgang, but not forcing $n^2/4$ arrows per direction:

Forcing diagonal symmetry loses optimality.

I also tried forcing horizontal and vertical symmetry, $r(50)=34$, and minimizing the number of D (hence U). The minimum is at most $192$:

With horizontal and vertical symmetry and $r(64)=44$, the minimum number of D (hence U) is $328$:

I solved the LP relaxation for $n=50$ with no restrictions and got objective value $34+6/7$. The resulting optimal solution is not unique. Here's a plot where the color is determined by a fractional value $\ge 1/2$ (gray if the largest fractional value is $< 1/2$):

Here's a heat map of the fractional assignments of cell to direction:

Here's a heat map of the number of arrows pointing to each cell for an optimal LP solution with no restrictions:

And here's a heat map of the number of arrows pointing to each cell for an optimal ILP solution with no restrictions: