jmc

Since $n$ has the same remainder whether divided by 6 or by 8, we can write that $n=6a+r=8b+r$, where $0\le r\le 5$. This implies that $3a=4b$, and so $a$ is a multiple of 4 and we can write $a=4k$ for some integer $k$. Since $100<n<200$, we see that $95<6a<200$, or $\frac{95}{24}<k<\frac{200}{24}$. Since $k$ is an integer, $4\le k\le 8$. If $k=4$, then we must have $r=5$. Otherwise, $r=0,1,2,3,4,5$ are all allowable. Thus we have a total of $\boxed{25}$ possible values for $n$.