Solution:

For each of the given lines, the number of intersection points lying off it is
$$m=\frac{(n-1)(n-2)}{2}$$
Suppose that, for every line, on one of its sides there are no more than $k$ intersection points, where $k \leqslant \frac{n}{2}$. Let us count in two ways the triples $(p, A, B)$, where $A$ and $B$ are intersection points on different sides of the line $p$. By the assumption, for each line $p$ there are no more than $k(m-k)$ such triples, so their total number is $N \leqslant n k(m-k)$.

Let us now consider two types of such triples:
type 1, when the points $A$ and $B$ lie on the same one of the given lines, and
type 2, when they do not.
Every quadruple of given lines, together with the intersection points they determine, determines exactly 4 triples of type 1, and every triple is determined by exactly one quadruple of lines. It follows that there are $4\binom{n}{4}$ triples of type 1. Also, every quintuple of given lines together with the intersection points they determine, determines at least 5 triples of type 2, so the number of triples of type 2 is at least $5\binom{n}{5}$. This can be checked directly: there are six different arrangements of five lines in general position, as in the figure.

Therefore, $N \geqslant 4\binom{n}{4}+5\binom{n}{5}=n\binom{n}{4}$.
It follows that $k(m-k) \geqslant\binom{n}{4}=\frac{m(m-1)}{6}$. From this one obtains
$$k \geqslant k_{0}=\frac{1}{2}\left(m-\sqrt{\frac{m^{2}+2 m}{3}}\right)$$
Since the inequality $k_{0}>\frac{m}{5}=\frac{(n-1)(n-2)}{10}$ reduces to $m>25$, which holds for $n \geqslant 9$, while for $n=8$ we have $k_{0}>\left[\frac{m}{5}\right]=4$, it remains to examine the cases $n \leqslant 7$.

For $n \leqslant 4$ the statement is trivial, as is the attainment of equality, since then $k_{0}=\left[\frac{m}{5}\right]=0$. For $n=5,6,7$ we have, respectively, $\left\lceil k_{0}\right\rceil=\left[\frac{m}{5}\right]=1,2,3$, so the statement of the problem holds in these cases as well, and equality is also attained in the cases shown in the figure.