24th Hellenic Mathematical Olympiad

We have $A_1{A}_2\dots A_{2007}\cdot B_1{B}_2\dots B_{2007}=1$, because each element of the table appears two times, one in a row and one in a column, and so the number of $(-1)$ in the product $A_1{A}_2\dots B_{2007}$ is even, say, for example $2k$.

Therefore the number of $+1$ will be $4014-2k$.

If $(4014-2k)(1)+2k(-1)=0\Rightarrow 4014=4k$, which is absurd, because $4$ does not divide $4004$. Hence $A_1+A_2+\cdots +A_{2007}+B_1+B_2+\cdots +B_{2007}\ne 0$.