Example 2 Arrange all fractions of the form $\frac{m}{n}\left(m, n \in \mathbf{N}_{+}\right)$ in a row according to the following rules:
(1) If $m_{1} n_{1}<m_{2} n_{2}$, then $\frac{m_{1}}{n_{1}}$ is placed before $\frac{m_{2}}{n_{2}}$;
(2) If $m_{1} n_{1}=m_{2} n_{2}$ and $n_{1}<n_{2}$, then $\frac{m_{1}}{n_{1}}$ is placed before $\frac{m_{2}}{n_{2}}$.
Therefore, the 12th number is $\qquad$, and $\frac{1990}{1}$ leads $\frac{1}{1990}$ by $\qquad$ numbers.
(1) For positive integers $m, n$, according to the rule, in $\frac{m}{n}$, $mn=1$, only $1 \times 1=1$; $mn=2$, only $1 \times 2=2$. Therefore, we can write down the sequence according to this rule:
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{1}{3}, \frac{4}{1}, \frac{2}{2}, \frac{1}{4}, \frac{5}{1}, \frac{1}{5}, \frac{6}{1}, \frac{3}{2}, \cdots
$$

Thus, the 12th number is $\frac{3}{2}$.
(2) Since $1990=1 \times 1990=2 \times 995$
$$
=5 \times 398=10 \times 199 \text{, }
$$

and each form corresponds to two fractions, there are 8 fractions where $mn=1990$, and $\frac{1990}{1}$ is the first, $\frac{1}{1990}$ is the last. Therefore, $\frac{1990}{1}$ leads $\frac{1}{1990}$ by 7 numbers.

Note: For simpler numbers, we can directly write them according to the rule; for more complex numbers, we can use the method of prime factorization to solve. For example, to find out how many numbers $\frac{2000}{1}$ leads $\frac{1}{2000}$ by: Since $2000=2^{4} \times 5^{3}$, there are $(4+1)(3+1)=20$ factors, so there are 20 fractions where $mn=2000$. Therefore, $\frac{2000}{1}$ leads $\frac{1}{2000}$ by 19 numbers.