Slovenian Mathematical Olympiad, 2008

If $n = 2008$ fill the first row of the board with $-1$ and all other squares with $1$. Then $a_k = 1$ for all $k$ and $b_l = -1$ for all $l$, so
$$a_1 + a_2 + \dots + a_{2008} + b_1 + b_2 + \dots + b_{2008} = 1 + \dots + 1 + (-1) + \dots + (-1) = 0.$$

Now, let $n = 2007$. We will show that we cannot choose the numbers so that
$$a_1 + a_2 + \dots + a_{2007} + b_1 + b_2 + \dots + b_{2007} = 0.$$
Assume, on the contrary, that this can be done. Obviously, each of $a_k$ and $b_l$ is either $1$ or $-1$. Let $t$ denote the number of negative elements of the set $\{a_1, \dots, a_{2007}\}$ and let $s$ be the number of negative elements in $\{b_1, \dots, b_{2007}\}$. Then
$$a_1 + \dots + a_{2007} = -t + (2007 - t) = 2007 - 2t$$
and
$$b_1 + \dots + b_{2007} = -s + (2007 - s) = 2007 - 2s.$$
This implies
$$0 = a_1 + a_2 + \dots + a_{2007} + b_1 + b_2 + \dots + b_{2007} = 2007 - 2t + 2007 - 2s,$$
$$\text{so } s + t = 2007.$$
How many $-1$s are there on the board? If $a_i = -1$, then the $i^{\text{th}}$ row contains an odd number of $-1$s. If $a_i = 1$, then this number is even. So, the parity of the number of $-1$s is the same as the parity of $t$. A similar argument for the columns shows that the parity of the total number of $-1$s is the same as the parity of $s$. So, $s$ and $t$ have the same parity, which contradicts the equation $s + t = 2007$ from above.

When $n = 2007$ we cannot fill the board with $1$s and $-1$s so that
$$a_1 + a_2 + \dots + a_{2007} + b_1 + b_2 + \dots + b_{2007} = 0.$$