jmc

Writing out the recursive statement for $a_n,a_{n-1},\dots ,a_{10}$ and summing them gives$$a_n+\cdots +a_{10}=100(a_{n-1}+\cdots +a_{10})+n+\cdots +10$$Which simplifies to$$a_n=99(a_{n-1}+\cdots +a_{10})+\frac{1}{2}(n+10)(n-9)$$Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12,23,34,45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20,31,42,53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{45}$. Note that we can also construct the solution using CRT by assuming either $11$ divides $n+10$ and $9$ divides $n-9$, or $9$ divides $n+10$ and $11$ divides $n-9$, and taking the smaller solution.