Parse $0 . \dot{a} b \dot{c}=\frac{\overline{a b c}}{999}$, where the numerator $\overline{a b c}<999.999=3^{3} \times 37$, and among the numbers from 1 to 999, there are 666 numbers not divisible by 3, among these 666 numbers, there are $\frac{999}{37}-\frac{999}{3 \times 37}=18$ numbers divisible by 37. There are also 12 fractions of the form $\frac{3 m}{37}(m=1,2, \cdots, 12)$ that can be simplified from $\frac{\overline{a b c}}{999}$. Therefore, the answer is $666-18+$ $12=660$ numbers.
19.2.15 ** Let $s$ be the set of all rational numbers $r$ that satisfy the following conditions: $0<r<1$, and $r$ can be expressed as a repeating decimal of the form $0 . a b c a b c a b c \cdots=0 . \dot{a} b \dot{c}$, where the digits $a, b, c$ do not necessarily have to be distinct. Write the elements of $s$ as reduced fractions. How many different numerators are there?