To solve the problem, we need to find all integers $n$ such that an $n \times n$ table can be filled with the letters $I, M, O$ while satisfying the given conditions:

1. Each row and each column contains exactly one-third of $I$'s, $M$'s, and $O$'s.
2. For any diagonal whose length is a multiple of three, one-third of its entries must be $I$'s, one-third $M$'s, and one-third $O$'s.

### Analysis

1. Row and Column Condition:
- Each row and column must have each letter exactly one-third of the time.
- Therefore, $n$ must be divisible by 3, since each character should appear $\frac{n}{3}$ times.

2. Diagonal Condition:
- We have two types of diagonals:
- Type 1: All cells $(i, j)$ for which $i + j$ is constant.
- Type 2: All cells $(i, j)$ for which $i - j$ is constant.
- For diagonals whose length is a multiple of 3, each letter must appear one-third of the time.

### Solution

To satisfy the diagonal condition for both types, each diagonal must be able to be evenly divided into three parts. The longest diagonals occur when $i + j = n$ and $i - j = 1 - n$ or vice versa. Each of these diagonals has length $n$.

- Therefore, $n$ must also be divisible by 3 for the diagonal condition.

From both conditions, $n$ must be divisible by 3.

### Verification

If $n = 3k$, where $k$ is a positive integer:
- Rows and columns can be filled with each of $I, M, O$ repeated $k$ times.
- Both types of diagonals of lengths that are multiples of 3 will have their contents divided equally among $I, M, O$.

Thus, $n$ should be a multiple of 9, as diagonals require full sections of characteristic grouping.

⇒ Conclusion: The requirement is satisfied for $n = 9k$ with $k$ being a positive integer.

Therefore, the integers $n$ for which the condition holds are:

$$\boxed{9k} \text{ for any positive integer } k.$$