Slovenija 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 $$a_1+a_2+\cdots +a_{2008}+b_1+b_2+\cdots +b_{2008}=1+\cdots +1+(-1)+\cdots +(-1)=0.$$

Now, let $n=2007$. We will show that we cannot choose the numbers so that $$a_1+a_2+\cdots +a_{2007}+b_1+b_2+\cdots +b_{2007}=0.$$ 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 $$a_1+\cdots +a_{2007}=-t+(2007-t)=2007-2t$$ and $$b_1+\cdots +b_{2007}=-s+(2007-s)=2007-2s.$$ This implies $$0=a_1+a_2+\cdots +a_{2007}+b_1+b_2+\cdots +b_{2007}=2007-2t+2007-2s,$$ $$\text{so }s+t=2007.$$ 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 $$a_1+a_2+\cdots +a_{2007}+b_1+b_2+\cdots +b_{2007}=0.$$