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 $2^3 \cdot 11 \cdot 23$. $\square$

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 $\frac{1}{4}n!(n-1)!n(n-1)(n-2) = \frac{1}{4}n!(n^2-3n+2)$.
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 $n! \cdot n$ 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 $a_9 = 9! \cdot 23$ is not divisible by $11$, and the next suitable $n$ is $13$, where $a_{13} = 13! \cdot 46$ is divisible by $2^3 \cdot 11 \cdot 23$. $\square$