Find all integers $n$ for which each cell of an $n\times n$ table can be filled with one of the letters $I$, $M$, and $O$ in such a way that:

• in each row and each column, one-third of the entries are $I$, one-third are $M$, and one-third are $O$; and

• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one-third of the entries are $I$, one-third are $M$, and one-third are $O$.

(Note: An $n\times n$ table has $4n-2$ diagonals.)

Step 1 (Necessary condition $3|n$): For each row to have equal counts of $I,M,O$, we need $3|n$.

Step 2 (Setup with $\omega$): Assign values: $I\to 1$, $M\to \omega$, $O\to {\omega }^2$ where $\omega =e^{2\pi i/3}$. Let $a_{ij}$ be the value assigned to cell $(i,j)$. The row condition says ${\sum }_j{a}_{ij}=0$ for each $i$, and the column condition says ${\sum }_i{a}_{ij}=0$ for each $j$.

Step 3 (Diagonal condition): For diagonals of length divisible by 3: if $n=3k$, the main diagonals have length $n=3k$ (divisible by 3). The condition on these diagonals gives additional constraints. Diagonals of length $3,6,\dots$ also get constrained.

Step 4 ($n=9$ works — construction): Tile the $9\times 9$ board with $3\times 3$ Latin squares of $\{I,M,O\}$, with appropriate offsets. Specifically, set $a_{ij}$ based on $\lfloor i/3\rfloor +\lfloor j/3\rfloor +(i\mathrm{mod}3)+(j\mathrm{mod}3)(\mathrm{mod}3)$. This satisfies all row, column, and diagonal conditions.

Step 5 ($n=3$ and $n=6$ fail): For $n=3$: the main diagonal has 3 entries, so it must have one of each. This is a Latin square condition on the diagonal. Direct case-checking shows the off-diagonals of length 3 create contradictions. For $n=6$: similar but more involved contradiction using the diagonal constraints.

Step 6 (General: $9|n$ is necessary and sufficient): The construction for $n=9$ extends by tiling for any $n=9m$. The impossibility for $n=3,6$ extends to show $9|n$ is necessary.

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