jmc

Since $0.\overline{ab}=\frac{ab}{99}$, the denominator must be a factor of $99=3^2\cdot 11$. The factors of $99$ are $1,$ $3,$ $9,$ $11,$ $33,$ and $99$. Since $a$ and $b$ are not both nine, the denominator cannot be $1$. By choosing $a$ and $b$ appropriately, we can make fractions with each of the other denominators.

Thus, the answer is $\boxed{5}$.