I found this Olympiad style combinatorics problem. Does anybody know how to solve it?
Let $n\geq2$ be a natural number. An $n\cdot n$ grid is drawn on a board and each field with one of the numbers $-1$ or $+1$ labeled. Then the $n$ row sums and also the $n$ column sums is calculated and the sum $S_n$ of all these $2n$ sums is determined.
(a) Show that for no odd number $n$ there is a label with $S_n = 0$.
(b) Show that if $n$ is an even number, then there are at least six different labels with $S_n = 0$.
The (a) part is easy. First, note that $S_n = 2 A_n$, where $A_n$ is the sum of all $n²$ numbers in the grid. Then, note that if $n$ is odd, then also $A_n$ must be odd. To see this, you can consider the invariant (an Olympiad concept) of the parity of $A_n$. Whatever the label is, when you consider $A_n$ you are adding up $n²$ odd numbers (+1 or -1), so the sum must be odd. Finally, if $A_n$ is odd it cannot be zero, so also $S_n$ cannot be 0.
The part (b) is constructive. You should find at least 6 labels for which $A_n$ is zero. Try starting from a 2x2 board.
$$\begin{bmatrix} +1 & -1 \\ +1 & -1 \end{bmatrix}, \ \begin{bmatrix} +1 & -1 \\ -1 & +1 \end{bmatrix}, \ \begin{bmatrix} +1 & +1 \\ -1 & -1 \end{bmatrix}$$
and so on...