Indeed it can be solved as a traveling salesman problem on 13 nodes, but brute force is not required. First define an undirected graph with 145 nodes, one per box, with an edge between each pair of horizontally or vertically adjacent boxes (unless both are gray). Define each edge weight to be the average of the numbers of its two incident nodes. Now solve a shortest path problem for each pair of start point, objective, or end point. The resulting shortest path distances $d_{i,j}$ are:
\begin{matrix}
i\backslash j &13 &20 &52 &65 &68 &76 &101 &106 &132 &142 &145\\
\hline
1 &43 &16 &38 &13 &27 &19 &26 &41 &31 &33 &43 \\
13 &&33 &5 &56 &16 &52 &35 &10 &39 &16 &13 \\
20 &&&28 &29 &17 &31 &26 &31 &31 &23 &33 \\
52 &&&&51 &11 &47 &30 &5 &34 &11 &8 \\
65 &&&&&40 &6 &22 &51 &19 &41 &51 \\
68 &&&&&&39 &22 &14 &26 &15 &17 \\
76 &&&&&&&22 &46 &13 &36 &46 \\
101 &&&&&&&&29 &11 &19 &29 \\
106 &&&&&&&&&33 &10 &5 \\
132 &&&&&&&&&&23 &33 \\
142 &&&&&&&&&&&10
\end{matrix}
Now add a dummy node 146 that is adjacent only to the start point 76 and end point 106, with $d_{76,146} = d_{106,146} = 0$. This is a standard trick to convert a TSP path problem with specified start and end points into a TSP tour problem. Finally, solve the TSP on these 13 nodes, yielding optimal tour with edges
$$(1,20),(20,68),(13,68),(13,52),(52,142),(142,145),(106,145),(106,146),(76,146),(76,132),(101,132),(65,101),(1,65)$$
and total cost
$139$, as found by @DanielMathias. Here is a plot, omitting the two dummy edges:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/ELiFV.png)