[Example 3.6.4] There are $h+s$ lines on a plane, where $h$ lines are horizontal, and $s$ lines satisfy:
(1) None of them are horizontal lines;
(2) No two of them are parallel;
(3) No three of the $h+s$ lines are concurrent, and these $h+s$ lines exactly divide the plane into 1992 regions.
Find all pairs of positive integers $(h, s)$.
Given that there are two types of lines, one being horizontal and the other non-horizontal, we can study their division of the plane in steps. Therefore, we can calculate in three steps:
Step one, consider the regions divided by $s$ lines;
Step two, see what changes occur when $h$ horizontal lines are added to the already divided plane;
Step three, solve for the specific 1992 regions.
First, solve step one: How many regions do $s$ lines divide the plane into, assuming no two lines are parallel and no three lines intersect at the same point?
One line divides the plane into two regions.
Let $n$ lines divide the plane into $a_{n}$ regions. Now consider what changes when one more line is added to make $n+1$ lines.
Since the added line is not parallel to the previous $n$ lines and no three lines intersect at the same point, this line is divided into $n+1$ segments (line segments or rays) by the original $n$ lines. Each segment divides an existing region into two, thus
$$
\begin{array}{c}
a_{n+1}=a_{n}+(n+1) . \\
a_{1}=2,
\end{array}
$$

Then,
$$
a_{n}=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)+a_{1}
$$
$$
\begin{array}{l}
=n+(n-1)+\cdots+2+2 \\
=1+\frac{n(n+1)}{2} .
\end{array}
$$

Thus, $s$ lines divide the plane into $\frac{s(s+1)}{2}+1$ regions.
Step two, add $h$ horizontal lines.
Since each added horizontal line is divided into $s+1$ parts (line segments or rays) by the original $s$ lines, these $h$ horizontal lines further divide the plane into $h(s+1)$ regions.
From steps one and two, $h+s$ lines divide the plane into
$$
h(s+1)+1+\frac{s(s+1)}{2}
$$

regions.
Step three: Solve for $(h, s)$ given 1992 regions.
$$
\begin{array}{c}
h(s+1)+1+\frac{s(s+1)}{2}=1992, \\
(s+1)(2 h+s)=2 \times 1991=2 \times 11 \times 181 .
\end{array}
$$

Thus, $s+1$ is a divisor of $2 \times 1991$.
Note that
$$
s^{2}<(s+1)(s+1)<(s+1)(s+2 h)=2 \times 11 \times 181,
$$

so
$$
s<\sqrt{2 \times 11 \times 181}<5 \times 14=70 .
$$

Therefore,
$$
s+1 \leqslant 70 \text {. }
$$

Thus, $s+1$ can only be $2, 11, 22$.
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{1}{|c|}{$s+1$} & \multicolumn{1}{c|}{$s$} & \multicolumn{1}{c|}{$2 h+s$} & \multicolumn{1}{c|}{$h$} \\
\hline 2 & 1 & 1991 & 995 \\
11 & 10 & 362 & 176 \\
22 & 21 & 181 & 80 \\
\hline
\end{tabular}

Therefore, the integer pairs $(h, s)$ that satisfy the problem are
$$
(h, s)=(995,1),(176,10),(80,21)
$$

with a total of three solutions.