Solution:

Because $100 \equiv 1 \pmod{99},$ the recurrence gives $a_n \equiv a_{n-1} + n \pmod{99},$ so $$a_n \equiv 10 + 11 + \cdots + n = \frac{(n + 10)(n - 9)}{2} \pmod{99}.$$ We need $99 \mid \frac{(n+10)(n-9)}{2}.$ One of $n + 10$ and $n - 9$ is even, so this is the same as requiring $9$ and $11$ each to divide the product. Since the two factors differ by $19,$ they cannot both be multiples of $3.$

So $9$ must divide one factor entirely and $11$ the other (or one factor is divisible by $99$). Checking the cases: $99 \mid n - 9$ first at $n = 108;$ $99 \mid n + 10$ first at $n = 89;$ $9 \mid n + 10$ with $11 \mid n - 9$ first at $n = 53;$ and $11 \mid n + 10$ with $9 \mid n - 9$ first at $n = 45.$

The least is $n = 45,$ where $\frac{55 \cdot 36}{2} = 990$ is indeed a multiple of $99.$