, 2015

Let us replace every number $n$ distributed into a box in the grid by $p = 0, 1, 2$ which satisfies the identity $n \equiv p \pmod 3$. The condition that the sum of the numbers on each row and on each column must be a multiple of $3$ does not change after the replacement of the numbers $n$ by $p$. Among the triples $(a, b, c)$ of integers lying in between (and including) $0$ and $2$, only $(0, 0, 0)$, $(1, 1, 1)$, $(2, 2, 2)$ and $(0, 1, 2)$ and its permutation of entries are the ones satisfying the requirement that the sum of the entries of the three numbers equals a multiple of $3$. This means that only possible triplets to satisfy the requirement of the problem after the replacement of $n$ by $p$ are either the ones with all the same entries (such as $(1, 1, 1)$) or the ones with all distinct entries (such as $(0, 1, 2)$).

If we represent the remainders $0$, $1$, $2$ by the letters $A$, $B$, $C$ (disregarding the order), we see that each of the letters $A$, $B$, $C$ can represent any of the three numbers $0$, $1$, $2$ and the only possible placement of these letters that can satisfy the requirements of the problem must be one of the four placements indicated below:




For each of the placements shown above, there are $3!$ ways of associating $3$ kinds of remainder to each of the three letters, and therefore, there are $4 \cdot 3! = 24$ possible ways of getting the distinct configurations after the replacement of numbers $n$ by remainders $p$. Furthermore, since there are $3$ possible ways of choosing integers $n$ for each of the possible remainders $p$, there are $(3!)^3 = 216$ ways of inserting integers for each of the $24$ possible configurations of the placements of remainders $p$ considered above.
Therefore, there are $24 \cdot 216 = 5184$ ways of distributing numbers into the boxes of the grid, so as to satisfy the requirement of the problem.