1.9 Let $s$ be the set of all rational numbers $r$ that satisfy the following conditions:
$$
\text { (1) } 0<r<1 \text {; }
$$
(2) $r=0 . a b c a b c a b c \cdots=0 . a b \dot{c}$, where $a, b, c$ are not necessarily distinct.

How many different numerators are there when the numbers $r$ in $s$ are written in their simplest form?
[Solution] Since $0 . \dot{a} b \dot{c}=\frac{a b c}{999}$, and because $999=3^{3} \times 37$, therefore, when the positive integer $a b c$ is neither divisible by 3 nor by 37, $\frac{a b c}{999}$ is in its simplest form, and the numerator is $a b c$. Among the 999 positive integers from 1 to 999, there are $\frac{999}{3}=333$ numbers divisible by 3; $\frac{999}{37}=27$ numbers divisible by 37; and 9 numbers divisible by $3 \times 37$. By the principle of inclusion and exclusion, there are $999-333-27+9=648$ such different numerators.

Next, consider which of the excluded numbers, either multiples of 3 or 37 or both, can still serve as the numerators of the simplest fractions.

Since $37^{2}>999$, any multiple of 37, when simplified by dividing by 37, will no longer be a multiple of 37. Therefore, any multiple of 37 cannot be the numerator of the simplest fraction.

However, among the multiples of 3, there are some that can be simplified by dividing by $3^{3}$ to produce new numerators, i.e., the following 12 numbers: $\square$
$$
\{81 k \mid k=1,2, \cdots, 12\}
$$

When simplified by dividing by 999, the simplest fractions obtained are
$$
\frac{3 k}{37}, k=1,2, \cdots, 12 \text {. }
$$

Their numerators are $3,6,9, \cdots, 36$. These are all multiples of 3 and are certainly different from the previous ones. Therefore, the total number of different numerators that meet the requirements is 660.