Note that $a_n=\binom{200}{n}\times 3^{\frac{200-n}{3}}\times 2^{\frac{400-5n}{6}}$.

In order for $a_n$ to be an integer, both $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ must be integers. The $2^{nd}$ term can be rewritten as $(36-n)+\frac{n+4}{6}$. Therefore, $(n+4)$ must be a multiple of $6$. When $6\mid (n+4)$, the $1^{st}$ term, $\frac{200-n}{3}$, will be an integer too.

When $n=6k+2$, ($k=0,1,\cdots, 13$), both $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ are non-negative integers. Therefore corresponding $a_n$ must be integer. There are totally 14 such qualified terms.

When $n=86$, $a_{86}=\binom{200}{86}\times3^{38}\times 2^{-5}$. It will be an integer if and only if the coefficient $\binom{200}{86}=\frac{200!}{86!\times 114!}$ contains at least five factors of $2$. The number of factor $2$ in $200!$ can be computed as: $$\left\lfloor{\frac{200}{2}}\right\rfloor+\left\lfloor{\frac{200}{2^2}}\right\rfloor+\cdots + \left\lfloor{\frac{200}{2^7}}\right\rfloor=198$$

Similarly, we find $86!$ contains $82$ factors of $2$, and $114!$ contains $110$. Therefore, the number of factor $2$ in $\binom{200}{86}\frac{200!}{86!\times 114!}$ is $197-82-110=5$. This means $a_{86}$ is an integer.

When $n=92$ (which is the largest number satisfying $6k+2$ in the given range), $a_{92}=\binom{200}{92}\times 2^{36}\times 2^{-10}$. It can be shown that $\binom{200}{92}$ does not contain $10$ or more $2$ in its factorization. So $a_{92}$ is not an integer.

In summary, there are $14+1=\boxed{15}$ integers in $\{a_n\}$.