Let $\mathcal{L}\subseteq\mathbb{Z}^n$ be an integer lattice (not necessarily full-rank). Is it guaranteed that there is a vector $\mathbf{v}\in\mathcal{L}$ such that $\|\mathbf{v}\|=\det(\mathcal{L})$?
If $\mathcal{L}$ is full-rank, then the answer is certainly yes. Let $\mathbf{B}$ be a matrix whose columns form a basis for $\mathcal{L}$. Then $\|\det(\mathbf{B})\|=\det(\mathcal{L})$. Then the adjugate of $\mathbf{B}$, namely $\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{B}^{-1}$, is a matrix of integer entries. Therefore, the columns of $\mathbf{B}\cdot\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{I}_n$ are in $\mathcal{L}$. But the columns are just $\det(\mathbf{B})$ times the standard basis vectors. Each of these scaled standard basis vectors has norm $\|\det(\mathbf{B})\|=\det(\mathcal{L})$.
What about integer lattices that are not full rank?
(Note that the requirement that $\mathcal{L}\subseteq \mathbb{Z}^n$ is an integer lattice is important here. For arbitrary real lattices $\mathcal{L}\subseteq\mathbb{R}^n$, the statement is false. For example, the lattice $\frac{1}{2}\cdot \mathbb{Z}^n$ has determinant $2^{-n}<1/2$ for $n\geq 2$. However all vectors have the norm at least $1/2$.)