$10 \cdot 120$ taking out the positive rational numbers less than 1 with a denominator of 1000, reduce each to its simplest form, and then add the numerator and denominator of each number. How many different numbers can be obtained.
(17th American Invitational Mathematics Examination, 1999)
[Solution] Since $1000=2^{3} \cdot 5^{3}$, 1000 has 16 divisors:
$1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000$.
All 16 numbers except 1 can serve as the denominators of fractions less than 1 with a denominator of 1000.

Classifying all rational numbers in the problem by their denominators results in 15 categories, with the category of denominator $k$ being called the $k$ category.

The sum of the numerator and denominator of a reduced fraction $-\frac{q}{p}$ is called the generating number of this fraction. Clearly,
the only fraction with a denominator of 2 is $\frac{1}{2}$, with a generating number of 3;
the fractions with a denominator of 4 are $\frac{1}{4}, \frac{3}{4}$, with generating numbers of 5 and 7;
the fractions with a denominator of 5 are $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$, with generating numbers of 6, 7, 8, and 9;
the fractions with a denominator of 8 are $\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}$, with generating numbers of 9, 11, 13, and 15.
Next, we prove that all odd numbers not less than 17 and not divisible by 5 and less than 2000 can be generated, while all odd numbers not less than 17 and less than 2000 that are divisible by 5 cannot be generated.
Let $M=\{m \mid 17 \leqslant m<2000, m \in N, 5 \nmid m, 2 \nmid m\}$
$$
=\{17,19,21,23,27,29,31,33,37, \cdots, 1999\} \text {. }
$$

For each $i \in M$, an even divisor $k$ of 1000 can be found such that $k < i < 2k$. Since $k$ has only the divisors 2 and 5, and since $2 \nmid i$ and $5 \nmid i$, it follows that $(k, i)=1$ and $(k, i-k)=1$, meaning $\frac{i-k}{k}$ belongs to the $k$ category and generates $i$.

Next, we prove that all odd numbers not less than 17 and less than 2000 that are divisible by 5 cannot be generated.
Clearly, the categories with denominators of $2,4,5,8$ cannot generate multiples of 5 between 17 and 2000.
The other categories have denominators $\{10,20,25, \cdots, 1000\}$, all of which are multiples of 5. Assuming $i (5 \mid i, 2 \nmid i, 17 \leqslant i<2000, i \in N)$ can be generated, it must belong to the category with denominator $k (5 \mid k)$, so $5 \mid i-k$ and $(k, i-k)=5$, meaning the fraction $\frac{i-k}{k}$ is not a reduced fraction. Therefore, $i$ cannot be generated.
Finally, we discuss the even numbers that can be generated.
Clearly, for categories with even denominators, the numerators can only be odd, so even numbers cannot be generated.
Thus, we only need to consider the cases with denominators of $5,25,125$.
In these cases, to generate even numbers, the numerators must be odd and not multiples of 5 and less than the denominator.
Since there are 999 odd numbers between 3 and 2000.
Among them, the numbers ending in 5 are $\frac{1995-15}{10}=198$.
For a denominator of 5, there are 2 possible numerators.
For a denominator of 25, there are $12-\left[\frac{12}{5}\right]=10$ possible numerators.

For a denominator of 125, there are $\frac{124}{2}-10-2=50$ possible numerators.
Therefore, a total of $999-198+2+10+50=863$ different numbers are obtained.