Bulgarian Winter Tournament

Exactly one of the colors will be used for two of the balls; let their numbers be $a$ and $b$, such that $a\le b$.

If $a>1$, then we have $(n-1)(n-2)/2$ choices for $a$ and $b$, and $n$ choices for their color. The remaining $n-1$ balls (two of which are the same) must be colored in the remaining $n-1$ colors; for this we have $(n-1)!/2$ variants. Therefore, in this case the number of possible colorings is $\frac{1}{4}(n-1)!.n(n-1)(n-2)=\frac{1}{4}n!(n^2-3n+2)$.

If $a=1$, then there are $n!$ options for coloring all balls except $a$ in $n$ colors. Now there are $n$ choices for the color of $a$. Thus, in this case there are $n!.n$ possible colorings.

Finally $a_n=\frac{1}{4}n!(n^2-3n+2+4n)=\frac{1}{4}n!(n^2+n+2)$.

If $a_n$ is a multiple of $2024=2^3\cdot 11\cdot 23$ and $n<23$, it must be $23|n^2+n+2$. Direct inspection shows that the smallest such $n$ is $9$, but $a_9=9!.23$ is not divisible by $11$, and the next suitable $n$ is $13$, where $a_{13}=13!.46$ is divisible by $2^3\cdot 11\cdot 23$. $\square$