[Example 3.3.7] In the plane, there are 100 lines. Can there be 1985 intersection points among them.
On a plane, 100 lines, if no two lines or multiple lines are parallel, and no three lines or multiple lines intersect at one point, that is, they intersect pairwise, and the intersection points do not coincide, at this time, the number of intersection points is the maximum, which is
$$
C_{100}^{2}=\frac{100 \times 99}{2}=4950
$$

intersection points.
The problem requires only 1985 intersection points, which means we need to reduce
$$
4950-1985=2965
$$

intersection points from the state where all lines intersect pairwise and the intersection points do not coincide. This requires adjusting the pairwise intersecting lines, i.e., making some lines parallel or some lines intersect at a common point.
Consider a bundle of lines intersecting at a common point.
Suppose there are $k$ such bundles, with the number of lines in each bundle being $n_{1}, n_{2}, \cdots$, $n_{k}$, where $n_{i} \geqslant 3, i=1,2, \cdots, k$.
Of course, the total number of lines in these $k$ bundles is
$$
n_{1}+n_{2}+\cdots+n_{k} \leqslant 100 .
$$

We know that for a bundle of $n_{1}$ lines intersecting at a common point, it can reduce $C_{n_{i}}^{2}-1$ intersection points.
Our goal is to reduce 2965 intersection points, which requires satisfying the equation
$$
\sum_{i=1}^{k}\left(C_{n_{i}}^{2}-1\right)=2965 .
$$

First, select the $C_{n_{1}}^{2}-1$ closest to 2965, i.e.,
$$
\begin{array}{l}
\frac{n_{1}\left(n_{1}-1\right)}{2}-1 \leqslant 2965, \\
n_{1}^{2}-n_{1}-5932 \leqslant 0, \\
\frac{1-\sqrt{23729}}{2} \leqslant n_{1} \leqslant \frac{1+\sqrt{23729}}{2} \text {, } \\
n_{1} \leqslant \frac{155}{2} \text {. } \\
n_{1} \leqslant 77 \text {. } \\
\end{array}
$$

That is,
a bundle of 77 lines intersecting at a common point can reduce $C_{77}^{2}-1=2925$ intersection points.
By $2965-2925=40$, our task is to find a bundle of lines intersecting at a common point that can reduce close to 40 intersection points, consider
$$
\begin{array}{c}
C_{n_{2}}^{2}-1 \leqslant 40 \\
n_{2}^{2}-n_{2}-82 \leqslant 0
\end{array}
$$

This yields
$$
n_{2} \leqslant 9 .
$$

And $C_{9}^{2}-1=35$, which reduces 35 intersection points.
Now, we need to reduce $40-35=5$ intersection points. Since $C_{4}^{2}-1=5$, a bundle of 4 lines intersecting at a common point reduces 5 intersection points, thus reducing a total of $2925+35+5=2965$ intersection points, meeting the problem's requirements.

From the above, after three adjustments, we can achieve a state of 100 lines, where 77 lines intersect at point $N_{1}$, 9 lines intersect at point $N_{2}$, 4 lines intersect at point $N_{3}$, and the remaining 10 lines intersect pairwise, at this time there are exactly
$$
C_{100}^{2}-\left(C_{77}^{2}-1\right)-\left(C_{9}^{2}-1\right)-\left(C_{4}^{2}-1\right)=1985
$$

intersection points.
Another method is
$$
\begin{aligned}
& C_{10}^{2}+C_{77}^{1} C_{9}^{1}+C_{77}^{1} C_{4}^{1}+C_{9}^{1} C_{4}^{1}+C_{77}^{1} C_{10}^{1}+C_{9}^{1} C_{10}^{1}+C_{4}^{1} C_{10}^{1}+3 \\
= & 45+693+308+36+770+90+40+3 \\
= & 1985 .
\end{aligned}
$$