jmc

Let $f(n)$ be the sum of the digits of $n$. It turns out that $n-f(n)$ is always divisible by 9. As a proof, write $n=a_{k}10^k+a_{k-1}10^{k-1}+\cdots +a_{1}10^1+a_0$. Therefore, $n-f(n)=a_k(10^k-1)+a_{k-1}(10^{k-1}-1)+\cdots +a_2(10^2-1)+a_1(10-1)$. Note that, in general, $10^n-1$ is divisible by 9 because $10^n-1$ is actually a string of $n$ 9's. Therefore, we can factor a 9 out of the right-hand side, so $n-f(n)$ is always divisible by 9. Note furthermore that $n-f(n)$ is always nonnegative, and that $f(n)$ and $n$ share the same remainder when divided by 9 (these are corollaries, the first coming from observation, the second being a direct result of the proof).

Now, consider $f(a_n)$, which is divisible by 9 if and only if $a_n$ is. We have $f(a_n)=f(1)+f(2)+\cdots +f(n-1)+f(n)$. Since $f(k)$ and $k$ have the same remainder when divided by 9, so we can substitute $k$ for $f(k)$ in each term without changing the remainder when divided by 9. Therefore, $f(a_k)\equiv \frac{k(k+1)}{2}(\mathrm{mod}9)$, which implies that we need either $k$ or $k+1$ to be divisible by 9. This happens either when $k$ is a multiple of 9 or when $k$ is one less than a multiple of 9. There are 11 multiples of 9 less than or equal to 100, and since 100 is not a multiple of 9, there are also 11 numbers which are one less than a multiple of 9 between 1 and 100. Therefore, there are $11+11=\boxed{22}$ values of $a_k$ that are divisible by 9 for $1\le k\le 100$.