, 2013

Solution:

Answer: $2013^{4025}$

We claim that the set of grids on which it is possible to obtain an array of all zeroes (mod $2013$) is indexed by ordered $4025$-tuples of residues (mod $2013$), corresponding to the starting entries in the first row and first column of the grid, giving the answer of $2013^{4025}$.

To do this, we show that after fixing all of the entries in the first row and column, there is a unique starting grid which can become an array of all zeroes after applying the appropriate operations.

Let $a_{i, j}$ be the entry in the $i$-th row and the $j$-th column. Suppose there is a sequence of operations giving all zeroes in the array; let $r_{i}$ be the number of times we operate on row $i$, and let $c_{j}$ be the number of times we operate on column $j$. It is enough to take all of these values to be residues modulo $2013$. Clearly, $a_{i, j} + r_{i} + c_{j} \equiv 0 \pmod{2013}$ for each $i, j$. In particular, $r_{1} + c_{1} \equiv a_{1,1}$.

Now, for each $i, j$, we have
$$\begin{aligned} a_{i, j} &\equiv -r_{i} - c_{j} \\ &\equiv (a_{i,1} + c_{1}) + (a_{1,j} + r_{1}) \\ &\equiv a_{i,1} + a_{1,j} - a_{1,1}, \end{aligned}$$
which is fixed. Thus, the rest of the entries in the grid are forced.

Conversely, if we set $a_{i, j}$ to be the appropriate representative of the residue class of $a_{i,1} + a_{1,j} - a_{1,1}$ modulo $2013$, we may take $r_{i} \equiv -a_{i,1} \pmod{2013}$, and $c_{j} \equiv a_{1,1} - a_{1,j} \pmod{2013}$ for each $i, j$. It is clear that $a_{i, j} + r_{i} + c_{j} \equiv 0 \pmod{2013}$ for each $i, j$, so we're done.