Iranian Mathematical Olympiad

Let $A_{i,j}$ denote the number in the cell located in the $i$-th row and $j$-th column of the table, and $S(i,j)$ the $3\times 3$ subtable whose rightmost upper number is $A_{i,j}$.

First, assume that $m,n\ge 3$ (If $m\le 2$ or $n\le 2$, there is no $3\times 3$ subtable). For $S(i,j)$, consider the following value $$M(i,j)=A_{i+1,j}-A_{i+2,j}+A_{i+2,j-1}-A_{i+1,j-2}+A_{i,j-2}-A_{i,j-1}.$$



Note that if we choose a row, a column or a diagonal of the table and add $1$ or $-1$ to the number in every cell of it, this value stays invariant.

Since we can make all the numbers in each subtable zero, we have $$M(i,j)=0\forall 1\le i\le n-2,3\le j\le m.$$

The assertion is proved by induction on $m+n$. If $m+n=6$, then $m=n=3$ and the assertion is trivial. If $m+n>6$, either $m>3$ or $n>3$. Without loss of generality we can assume that $m>3$. By omitting the last column, we get a $n\times (m-1)$ table. By the induction hypothesis, we can make the numbers in the cells of this $n\times (m-1)$ subtable zero, using allowable operations (except for the last column and the diagonal $A_{1,m}$ of the initial table).

Now, the numbers in the last column of the table may be nonzero. Considering the invariant values $M(i,m)$ ($1\le i\le n-2$) implies that $A_{i+1,m}=A_{i+2,m}$ for $1\le i\le n-2$. So $A_{2,m}=A_{3,m}=\cdots =A_{n,m}$. By using the last column several times all the numbers in the table except $A_{1,m}$ may become zero. But $A_{1,m}$ is itself a diagonal, hence we can also make this number zero.

If $m+n<6$, where either $m\le 2$ or $n\le 2$, the proof is very easy and is left to the reader.