Problem 1896

We first show that such a table exists when $n$ is a multiple of $9$. Consider the following $9 \times 9$ table.
$$\left(\begin{array}{ccccccccc} I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \end{array}\right)$$
It is a direct checking that the table (1) satisfies the requirements. For $n=9k$ where $k$ is a positive integer, we form an $n \times n$ table using $k \times k$ copies of (1). For each row and each column of the table of size $n$, since there are three $I$'s, three $M$'s and three $O$'s for any nine consecutive entries, the numbers of $I$, $M$ and $O$ are equal. In addition, every diagonal of the large table whose number of entries is divisible by $3$ intersects each copy of (1) at a diagonal with number of entries divisible by $3$ (possibly zero). Therefore, every such diagonal also contains the same number of $I$, $M$ and $O$.

Next, consider any $n \times n$ table for which the requirements can be met. As the number of entries of each row should be a multiple of $3$, we let $n=3k$ where $k$ is a positive integer. We divide the whole table into $k \times k$ copies of $3 \times 3$ blocks. We call the entry at the centre of such a $3 \times 3$ square a vital entry. We also call any row, column or diagonal that contains at least one vital entry a vital line. We compute the number of pairs ($l$, $c$) where $l$ is a vital line and $c$ is an entry belonging to $l$ that contains the letter $M$. We let this number be $N$.

On the one hand, since each vital line contains the same number of $I$, $M$ and $O$, it is obvious that each vital row and each vital column contain $k$ occurrences of $M$. For vital diagonals in either direction, we count there are exactly
$$1+2+\cdots+(k-1)+k+(k-1)+\cdots+2+1=k^{2}$$
occurrences of $M$. Therefore, we have $N=4k^{2}$.

On the other hand, there are $3k^{2}$ occurrences of $M$ in the whole table. Note that each entry belongs to exactly $1$ or $4$ vital lines. Therefore, $N$ must be congruent to $3k^{2} \bmod 3$.

From the double counting, we get $4k^{2} \equiv 3k^{2} (\bmod 3)$, which forces $k$ to be a multiple of $3$. Therefore, $n$ has to be a multiple of $9$ and the proof is complete.