Example 7 Given a set $S$ consisting of $2 n+1$ different irrational numbers, find the maximum positive integer $m$, such that there exist $m$ different numbers in $S$, the sum of any two of which is still an irrational number.
Let $S_{0}=\left\{ \pm \sqrt{2}, \pm \sqrt{5}, \pm \sqrt{10}, \cdots, \pm \sqrt{n^{2}+1}, \sqrt{(n+1)^{2}+1}\right\}$. Then $S_{0}$ contains $2 n+1$ irrational numbers, and among any $n+2$ of these numbers, there must be two that are opposites, whose sum equals 0 (a rational number). Therefore, the maximum value of $m$ is $\leqslant n+1$.

Let $S$ be a set of any $2 n+1$ distinct irrational numbers. Represent the numbers in $S$ as points in a plane (the numbers and their corresponding points are denoted by the same letter). If the sum of two numbers is a rational number, connect the corresponding two points with an edge; otherwise, do not connect them. This results in a graph $G$ with vertex set $V$.

First, we prove that $G$ contains no odd cycles. Assume $G$ has an odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Then $A_{1}+A_{2}, A_{2}+A_{3}, \cdots, A_{2 k}+A_{2 k+1}, A_{2 k+1}+A_{1}$ are all rational numbers. Their sum divided by 2 is $A_{1}+A_{2}+\cdots+A_{2 k+1}$. Thus, $A_{1}+A_{2}+\cdots+A_{2 k+1}$ is still a rational number, and
$$
A_{2 k+1}=\left(A_{1}+A_{2}+\cdots+A_{2 k+1}\right)-\left(A_{1}+A_{2}\right)-\left(A_{3}+A_{4}\right)-\cdots-\left(A_{2 k-1}+A_{2 k}\right)
$$
is a rational number, which is a contradiction. Therefore, $G$ contains no odd cycles.
By Theorem 4, $G$ is a bipartite graph $G=G(X, Y ; E)$. Thus, $|X|+|Y|=|V|=2 n+1$. Without loss of generality, assume $|X| \geqslant|Y|$, then $|X| \geqslant n+1$ and no two points in $X$ are adjacent, meaning that the sum of any two of the $n+1$ irrational numbers corresponding to the $n+1$ points in $X$ is an irrational number. Therefore, the maximum value of $m$ is $\geqslant n+1$.
In conclusion, the maximum value of $m$ is $n+1$.