Example 4 Given $n$ points in the plane with no three points collinear, $m$ line segments are drawn with these $n$ points as endpoints. It is known that for any two points $A$ and $B$ among these $n$ points, there is a point $C$ such that $C$ is connected to both $A$ and $B$ by line segments. Find the minimum value of $m$.
Let the $n$ points be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. First, consider an example.
If $n$ is odd, connect the line segments $A_{1} A_{2}, A_{1} A_{3}, \cdots$,
$$
A_{1} A_{n} ; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-1} A_{n} \text {. }
$$

If $n$ is even, connect the line segments $A_{1} A_{2}, A_{1} A_{3}, \cdots$,
$$
A_{1} A_{n} ; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-2} A_{n-1}, A_{2} A_{n} \text {. }
$$

Clearly, the line segments connected in the above manner satisfy the conditions. Therefore, the minimum value of $m$ is less than or equal to $\left\lfloor\frac{3 n-3}{2}\right\rfloor$. (Note: $\left\lfloor\frac{3 n-3}{2}\right\rfloor$ is the number of line segments connected in the above manner, where $\lfloor x\rfloor$ denotes the smallest integer not less than $x$.)

Next, we prove that these $n$ points require at least $\left\lfloor\frac{3 n-3}{2}\right\rfloor$ line segments to meet the requirements of the problem.

In fact, if each of the points $A_{1}, A_{2}, \cdots, A_{n}$ is connected by at least 3 line segments, then $m \geqslant \frac{3 n}{2}>\left\lfloor\frac{3 n-3}{2}\right\rfloor$;

If one of the points (let's say $A_{1}$) is connected by no more than 2 line segments, there are two cases:
(1) $A_{1}$ is connected by only 1 line segment, say $A_{1} A_{2}$. Then there is no point that is connected to both $A_{1}$ and $A_{2}$, which is a contradiction. Similarly, if $A_{1}$ is not connected by any line segment, a contradiction can also be derived.
(2) $A_{1}$ is connected by exactly 2 line segments $A_{1} A_{2}$ and $A_{1} A_{3}$. In this case, the line segment $A_{2} A_{3}$ must be connected (because for the two points $A_{1}$ and $A_{2}$, only $A_{3}$ can be connected to both $A_{1}$ and $A_{2}$).

Considering the points $A_{1}$ and $A_{i} (i \geqslant 4)$, $A_{i}$ must be connected to either $A_{2}$ or $A_{3}$, and by (1), $A_{i}$ must be connected by at least 2 line segments. Therefore, the points $A_{4}, A_{5}, \cdots, A_{n}$ (a total of $n-3$ points) must be connected by at least $\left(n-3+\left\lfloor\frac{n-3}{2}\right\rfloor\right)$ line segments. Thus, the number of line segments connected by these $n$ points is no less than
$$
3+(n-3)+\left\lfloor\frac{n-3}{2}\right\rfloor=\left\lfloor\frac{3 n-3}{2}\right\rfloor \text {. }
$$

Therefore, the minimum value of $m$ is $\left\lfloor\frac{3 n-3}{2}\right\rfloor$.