I think the following theorem is true but I'm not sure
Let $(A_{ij})$ be a nonzero $m\times n$ matrix of real numbers such that the sum of the entries of each row and each column is $0$. Then there are indices $i_1,j_1,i_2,j_2$ with $i_1\neq i_2$ and $j_1\neq j_2$ such that \begin{equation} A_{i_1j_1}>0, A_{i_2j_2}>0, A_{i_1j_2}<0 \text{ and } A_{i_2j_1}<0 \end{equation}
I'm looking for a proof or counterexample.
Edit: This "answer" is NOT true. See the comment section for a counterexample.
Suppose there is a counterexample $A$. By assumption, $A$ has a positive entry. WLOG, say $a_{11} > 0$. Since the sum of the first row and the first column are both $0$, there exist $i,j$ such that $a_{i1},a_{j1} < 0$. WLOG, say $i=j=2$. Then we have $$A = \begin{pmatrix}+ &- & \cdots \\ - & ?& \\ \vdots & & \ddots \\& & & &\end{pmatrix}$$ Notice that $a_{22} < 0$ $\color{red}{\textrm{(it should be } a_{22} \leq 0)}$ since $a_{11} > 0$ and $a_{12},a_{21} < 0$. Since the sum of the $2$nd row and the $2$nd column are both $0$, they consist positive entries, respectively. WLOG, say $a_{23},a_{32} > 0$ Hence $$A = \begin{pmatrix}+ &- & ?&\cdots \\ - & -&+ & \\ ?& + & ?& &\\ \vdots & & & \ddots \\& & & &\end{pmatrix}$$ Consider $a_{11},a_{32} > 0$ and $a_{12} < 0$, we get $a_{31} > 0$. Similarly, we get $a_{13} > 0$ since $a_{11}, a_{23} > 0$ and $a_{21} < 0$; $a_{33} > 0$ since $a_{23}, a_{32} > 0$ and $a_{22} < 0$. That is, $$A = \begin{pmatrix}+ &- & +&\cdots \\ - & -&+ & \\ +& + & +& &\\ \vdots & & & \ddots \\& & & &\end{pmatrix}$$ You can see that there's a pattern. In fact, it goes on forever! This implies that such $A$ doesn't exist.
