As each number of the grid appears exactly once in the sum of all columns of the grid and the same holds for the sum of all rows, we get that $S$ is twice the sum of all labels of the boxes of the $n \times n$ grid. Therefore, $S = 0$ holds if and only if the sum of all labels of the boxes vanishes, or equivalently, if the number of labels $+1$ equals the number of labels $-1$. We call such a labelling admissible.

a. If $n$ is odd, the sum of all labels is also odd, because it is a sum of an odd number of odd labels. Thus there cannot be an admissible labelling in this case.

b. If $n$ is even, we write $n = 2k$ for some integer $k$. The admissible labellings can be constructed as follows: Choose exactly half of the $n^2 = 4k^2$ boxes arbitrarily and label each of them with $+1$. The remaining boxes are labelled with $-1$.
Thus there are exactly $a_k := \binom{4k^2}{2k^2}$ admissible labellings of a $2k \times 2k$ grid.
We have $a_1 = \binom{4}{2} = 6$ and it is easily seen that $a_k$ is increasing in $k$: if $1 \le k' < k$, each admissible labelling of any $2k' \times 2k'$ subgrid can be extended to an admissible labelling of the $2k \times 2k$ grid by choosing half of the extra $4k^2 - 4k'^2$ boxes and labelling each of them with $+1$ and the remaining boxes with $-1$. Therefore, $a_k \ge 6$ for all $k$.