1. (40 points) If the simplest fraction $\frac{p}{q}$ is written in decimal form as $0 . a b a b a b a b \cdots$ (non-negative integers $a$ and $b$ can be equal, but at least one is non-zero), then, among the fractions that meet the condition, how many different numerators are there?
1. Decimal numbers in the form of 0.abababab $\cdots$ can all be written as $\frac{k}{99}$ $\left(k \in \mathbf{N}_{+}, k=1,2, \cdots, 98\right)$. Among the numbers $1,2, \cdots$, 99, those that are multiples of 3 or 11 are
$$
\left[\frac{99}{3}\right]+\left[\frac{99}{11}\right]-\left[\frac{99}{3 \times 11}\right]=39 \text {. }
$$

Therefore, the number of positive integers in $1,2, \cdots, 99$ that are coprime with 99 is $99-39=60$.

Additionally, there are 3 multiples of 27 (since $\frac{27}{99}=\frac{3}{11}, \frac{54}{99}=\frac{6}{11}$, $\frac{81}{99}=\frac{9}{11}$, we need to add back the numbers $3, 6, 9$), so the total number of different numerators is $60+3=63$.