Let $P$ be a convex $n$-polygon where each side and diagonal is colored with one of $n$ distinct colors. We need to determine for which $n$ there exists a coloring method such that for any three of the $n$ colors, we can always find one triangle whose vertices are vertices of $P$ and whose sides are colored by the three colors respectively.

First, we observe that $n$ cannot be even. This is because if any two distinct triangles get different color triples, there must be exactly $\binom{n-1}{2}$ triangles that get color $i$, and since every segment is part of exactly $n-2$ triangles, there must be $\frac{n-1}{2}$ segments of color $i$. This is an integer only if $n$ is odd.

For odd $n$, consider the following coloring method: label the vertices with $1, \ldots, n$ and color segment $ij$ by the $(i + j \mod n)$-th color.

To verify that this works, assume for two triangles $i_1, j_1, k_1$ and $i_2, j_2, k_2$ we have $i_1 + j_1 \equiv i_2 + j_2 \mod n$, $j_1 + k_1 \equiv j_2 + k_2 \mod n$, and $k_1 + i_1 \equiv k_2 + i_2 \mod n$. Then we would have:
$$(i_1 + j_1) + (j_1 + k_1) - (i_1 + k_1) \equiv (i_2 + j_2) + (j_2 + k_2) - (i_2 + k_2) \mod n,$$
which implies $n \mid 2(j_1 - j_2)$. Since $n$ is odd, we also have $n \mid j_1 - j_2$, and as $|j_1 - j_2| < n$, it must be that $j_1 = j_2$. Similarly, $i_1 = i_2$ and $k_1 = k_2$, so the triangles were not different after all and the coloring is as desired.

Therefore, the answer is: $\boxed{n \text{}}$ must be odd}}.