jmc

To have 9 as a factor, $n!$ must have two factors of 3. The first such $n$ for which this is true is 6, since $6!=\text{6}\times 5\times 4\times \text{3}\times 2\times 1$. Since 9 is a factor of $6!$ and $6!$ is a factor of $n!$ for all $n\ge 6$, the numbers $6!,7!,8!,\dots ,99!,100!$ are all divisible by 9. There are $100-6+1=\boxed{95}$ numbers in that list.