Austria2019

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 $a_k:=\left(\genfrac{}{}{0ex}{}{4k^2}{2k^2}\right)$ admissible labellings of a $2k\times 2k$ grid. We have $a_1=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ and it is easily seen that $a_k$ is increasing in $k$: if $1\le k^{\prime }<k$, each admissible labelling of any $2k^{\prime }\times 2k^{\prime }$ subgrid can be extended to an admissible labelling of the $2k\times 2k$ grid by choosing half of the extra $4k^2-4k^{\prime 2}$ boxes and labelling each of them with $+1$ and the remaining boxes with $-1$. Therefore, $a_k\ge 6$ for all $k$.