Answer: 45

Small Hint:

Since $100 \equiv 1 \pmod{99},$ the recurrence collapses to $a_n \equiv 10 + 11 + \cdots + n$ $\pmod{99}$

Big Hint:

You need $99$ to divide $\frac{(n+10)(n-9)}{2}.$ The factors differ by $19,$ so $3$ cannot divide both; split cases by which factor $9$ and $11$ divide.

Solution:

Because $100 \equiv 1 \pmod{99},$ the recurrence gives $a_n \equiv a_{n-1} + n \pmod{99},$ so $$\begin{aligned} &a_n \equiv 10 + 11 + \cdots + n \\ &\quad \small = \frac{(n + 10)(n - 9)}{2} \pmod{99}. \end{aligned}$$ We need $99$ to divide $\frac{(n+10)(n-9)}{2}.$ 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: $99$ divides $n - 9$ first at $n = 108;$ $99$ divides $n + 10$ first at $n = 89;$ $9$ divides $n + 10$ with $11$ dividing $n - 9$ first at $n = 53;$ and $11$ divides $n + 10$ with $9$ dividing $n - 9$ first at $n = 45.$

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