Example 2 Suppose there are $2 n(n \geqslant 2)$ points in space, where no four points are coplanar. Connect any $N$ line segments between them, such that these line segments form at least one triangle. Find the minimum value of $N$.


Thought analysis: By constructing examples, to show that $N \geqslant n^{2}+1$, and then prove that when $N=n^{2}+1$, if there are $N$ line segments connecting $2n$ points, then these line segments must form at least one triangle. The proof process can use mathematical induction, proof by contradiction, or the extreme principle. The proof process requires meticulous calculation.

Solution 1: Divide the $2n$ known points into two groups $S$ and $T$: $\square$
$$
S=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}, T=\left\{B_{1}, B_{2}, \cdots B_{n}\right\}.
$$

Now connect each pair of points $A_{i}$ and $B_{j}$ with a line segment $A_{i} B_{i}$, and do not connect any two points within the same group. There are a total of $n^{2}$ line segments. At this point, any three points among the $2n$ known points must have at least two points belonging to the same group, and these two points are not connected. Therefore, these $n^{2}$ line segments cannot form any triangle. This means that the minimum value of $N$ must be greater than $n^{2}$.

Next, we use mathematical induction to prove: If there are $n^{2}+1$ line segments connecting $2n$ known points, then these line segments must form at least one triangle.

When $n=2$, $n^{2}+1=5$, i.e., there are 5 line segments among four points. Clearly, these 5 line segments form two triangles. Assume that the proposition holds when $n=k (k \geqslant 2)$. When $n=k+1$, take any line segment $AB$. If the total number of line segments drawn from points $A$ and $B$ to the remaining $2k$ points is not less than $2k+1$, then there must be a point $C$ that is connected to both $A$ and $B$, thus forming $\triangle ABC$. If the total number of line segments drawn from points $A$ and $B$ is not more than $2k$, then after removing points $A$ and $B$, there are at least $k^{2}+1$ line segments among the remaining $2k$ points. By the induction hypothesis, they must form at least one triangle, thus completing the induction proof.
In summary, the minimum value of $N$ is $n^{2}+1$.
Solution 2: Constructing examples as in Solution 1, we know that the minimum value of $N$ is not less than $n^{2}+1$.

Since there are $n^{2}+1$ line segments among $2n$ points, on average, each point draws more than $n$ line segments. Therefore, we can conjecture that there must be a line segment whose two endpoints draw a total of at least $2n+1$ line segments. Let's prove this by contradiction.

Assume that the number of line segments drawn from $A_{1}, A_{2}, \cdots, A_{2n}$ are $a_{1}, a_{2}, \cdots, a_{2n}$, and for any line segment $A_{i} A_{j}$, $a_{i} + a_{j} \leqslant 2n$. Thus, the total number of line segments drawn from the two endpoints of all line segments does not exceed $2n(n^{2}+1)$. However, in this count, point $A_{i}$ is counted $a_{i}$ times, so we have
$$
\sum_{i=1}^{2n} a_{i}^{2} \leqslant 2n(n^{2}+1).
$$

On the other hand, it is clear that $\sum_{i=1}^{2n} a_{i}=2(n^{2}+1)$. By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{2n} a_{i}\right)^{2} \leqslant 2n\left(\sum_{i=1}^{2n} a_{i}^{2}\right), \\
\sum_{i=1}^{2n} a_{i}^{2} \geqslant \frac{1}{2n} \times 4(n^{2}+1)^{2} > 2n(n^{2}+1).
\end{array}
$$
(2) contradicts (1). Therefore, there must be a line segment whose two endpoints draw a total of at least $2n+1$ line segments. Let $A_{1} A_{2}$ be such a line segment, and there must be a point $A_{k} (k \geqslant 3)$ such that the line segments $A_{1} A_{k}$ and $A_{2} A_{k}$ both exist, thus forming $\triangle A_{1} A_{2} A_{k}$.

Solution 3: Constructing examples as in Solution 1, we know that the minimum value of $N$ is not less than $n^{2}+1$. Below, we use the extreme principle to prove that when $N=n^{2}+1$, these line segments must form at least one triangle. Therefore, the minimum value of $N$ is $n^{2}+1$.

Assume that there are $n^{2}+1$ line segments among $2n$ known points, and these line segments do not form any triangle. Let $A$ be the point among the $2n$ points that draws the most line segments, a total of $k$ line segments: $A B_{j}, j=1,2, \cdots, k$. Thus, any two points among $\left\{B_{1}, \cdots, B_{k}\right\}$ are not connected, otherwise, a triangle would be formed. Therefore, the number of line segments drawn from any $B_{j}$ does not exceed $2n-k$. Besides $A, B_{1}, \cdots, B_{k}$, there are $2n-k-1$ points, and the number of line segments drawn from any of these points does not exceed $k$. Thus, we have
$$
n^{2}+1 \leqslant \frac{1}{2}[k+k(2n-k)+(2n-k-1)k]
$$
$=k(2n-k) \leqslant n^{2}$, which is a contradiction. This completes the entire proof.