Let $a_1 = 24$ and form the sequence $a_n$, for $n \geq 2$, by the recurrence relation $a_n = 100a_{n-1} + 134$. The first few terms of the sequence are:
\[
24, 2534, 253534, 25353534, \ldots
\]
Determine the smallest value of $n$ for which $a_n$ is divisible by $99$.

Let's think step by step and output the final answer within \boxed{}.
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