Example 9 Consider $n^{2}$ points in the plane, where no three points are collinear, and $n \geqslant 4$ is a positive integer of the form $3 k+1$. Find the minimum number of line segments connecting these points such that for any $n$ points, we can find four points, each pair of which is connected by a line segment.
Lemma: Given $n(n \geqslant 2)$ points on a plane, some of them are connected by line segments, and there are no $p(p \geqslant 2)$ points such that every pair of them is connected. Let $B$ be the total number of line segments. Then $B \leqslant \frac{p-2}{2(p-1)} n^{2}$.

Proof of Lemma: When $p=2$, it is obvious. Assume the lemma holds for all $m, 2 \leqslant m \leqslant p(p \geqslant 2)$. Then for $p+1$, for any graph with $n$ points, there are no $p+1$ points such that every pair of them is connected. Let $A$ be the point with the most line segments, and let it have $l$ line segments. By the induction hypothesis, we have:
$$
B \leqslant \frac{p-2}{2(p-1)} l^{2} + l(n-l) = 
l\left(n - \frac{lp}{2(p-1)}\right) \\
= f(l) \leqslant f\left(\frac{p-1}{p} n\right) = \frac{p-1}{2p} n^{2},
$$

Therefore, the lemma also holds for $p+1$. By the principle of induction, the lemma is proved.

Since $n=3k+1$, we have $n^{2}=9k^{2}+6k+1=(k-1)(9k+6)+(9k+7)$.

Divide the $n^{2}$ points into $k-1$ groups of $9k+6$ points and one group of $9k+7$ points.

Consider connecting every pair of points within each group and not connecting points from different groups. Then the total number of line segments is $B=(k-1) \binom{9k+6}{2} + \binom{9k+7}{2} = \frac{(9k+6)(9k^2+5k+2)}{2}$.

Now, consider any graph that satisfies the problem's conditions, and consider the following two cases:
(1) There exist $2k+1$ points such that there is no triangle among them.

In this case, for any $k$ points among the remaining $9k^2+4k$ points, there is at least one line segment, i.e., there do not exist $k$ points such that no two of them are connected.
$$
\begin{array}{l}
\therefore B \geqslant \binom{9k^2+4k}{2} - \frac{k-2}{2(k-1)}(9k^2+4k)^2 = \\
\frac{(9k^2+4k)(9k^2+3k+1)}{2(k-1)},
\end{array}
$$

It is easy to verify that $\frac{(9k^2+4k)(9k^2+3k+1)}{2(k-1)} > \frac{(9k+6)(9k^2+5k+2)}{2}$.
(2) For any $2k+1$ points, there is a triangle among them.
(1) If there exist $k+1$ points such that no two of them are connected, then for any $k$ points among the remaining $9k^2+5k$ points, there is at least one line segment. As above, the total number of line segments will be more than $\frac{(9k+6)(9k^2+5k+2)}{2}$.
(2) If for any $k+1$ points, there is at least one line segment, by the lemma we have
$$
\begin{aligned}
& B \geqslant \binom{9k^2+6k+1}{2} - \frac{k-1}{2k}[k(9k+6)+1]^2 \\
= & \frac{(9k+6)(9k^2+5k+2)}{2} - \frac{k-1}{2k}.
\end{aligned}
$$

Since $B$ is an integer and $\frac{(9k+6)(9k^2+5k+2)}{2}$ is an integer, we have $B \geqslant \frac{(9k+6)(9k^2+5k+2)}{2}$.
In summary, the minimum number of line segments is:
$$
\frac{(9k+6)(9k^2+5k+2)}{2} = \frac{(n+1)(3n^2-n+4)}{2}.
$$