jmc

The prime factorization of 99 is $3^2\cdot 11$. So for $\frac{n}{99}$ to be in lowest terms, $n$ cannot be divisible by 3 or 11. The possible values of $n$ are from 1 to 98, inclusive, so there are 98 possible values for $n$. We can find the number of multiples of 3 and multiples of 11 and subtract from 98 to get the number of values that aren't divisible by 3 or 11. For multiples of 3, we go from 3 to 96, or $3\cdot 1$ to $3\cdot 32$, so there are 32 multiples of 3 from 1 to 98, inclusive. For multiples of 11, we go from 11 to 88, so there are 8 multiples of 11 from 1 to 98, inclusive. We have to make sure we don't double-count the numbers that are multiples of both 3 and 11: 33 and 66. So there are $32+8-2=38$ values of $n$ that are divisible by 3 or 11. That means there are $98-38=\boxed{60}$ values of $n$ for which $\frac{n}{99}$ is in lowest terms.