We have $a_n = C_{200}^n \cdot 3^{\frac{200-n}{3}} \cdot 2^{\frac{400-5n}{6}}$. When $a_n$ $(1 \le n \le 95)$ is an integer, $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ must be integers. Then $6 \mid n+4$.

When $n = 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80$, $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ are all non-negative integers. So the corresponding $a_n$, totally 14, are integers.

When $n = 86$, we have $a_{86} = C_{200}^{86} \cdot 3^{38} \cdot 2^{-5}$. The number of the factors of 2 in $200!$ is
$$\left[ \frac{200}{2} \right] + \left[ \frac{200}{2^2} \right] + \left[ \frac{200}{2^3} \right] + \left[ \frac{200}{2^4} \right] + \left[ \frac{200}{2^5} \right] + \left[ \frac{200}{2^6} \right] + \left[ \frac{200}{2^7} \right] = 197.$$
By the same reason, the numbers of the factors of 2 in $86!$ and $114!$ are 82 and 110, respectively. Therefore, the number of the factors of 2 in $C_{200}^{86} = \frac{200!}{86! \cdot 114!}$ is $197 - 82 - 110 = 5$. So $a_{86}$ is an integer.

When $n = 92$, we have $a_{92} = C_{200}^{92} \cdot 3^{36} \cdot 2^{-10}$. In the same way, we find the numbers of the factors of 2 in $92!$ and $108!$ are 88 and 105, respectively, which means that in $C_{200}^{92}$ is $197 - 88 - 105 = 4$. Therefore, $a_{92}$ is not an integer.

Overall, the required number is $14 + 1 = 15$. ☐