, 2022

The largest non-$n$-colorable number is $m_{\text{max}} = n(n-1) - 1$.

Suppose $m = m_{\text{max}}$. Then there must exist some color $C$ that appears on at most $n-2$ vertices, and these vertices divide all the vertices not colored $C$ into at most $n-2$ consecutive segments. Since there are at least $n(n-1) - 1 - (n-2) = (n-1)^2 > n(n-2)$ vertices not colored $C$, there must exist a consecutive segment of non-$C$-colored vertices of length at least $n$, and thus $m_{\text{max}}$ is a non-$n$-colorable number.

Now suppose $m \ge m_{\text{max}} + 1 = n(n-1)$. By the division algorithm, we have $m = nk + j$, where $k \ge n-1$ and $0 \le j \le n-1$. Then, starting from some vertex, going in clockwise order, we can first color in the sequence $[1, 2, \dots, n]$ repeated $k-j$ times, and then color in the sequence $[1, 1, 2, \dots, n]$ repeated $j$ times. It is easy to verify that this coloring indeed satisfies the requirements of the problem, so all $m > m_{\text{max}}$ are $n$-colorable numbers.