Example 1 If a set does not contain three numbers $x, y, z$ that satisfy $x+y=z$, then it is called simple. Let $M=\{1,2, \cdots, 2 n+1\}, A$ be a simple subset of $M$, find the maximum value of $|A|$. (1982
Let $A=\{1,3,5, \cdots, 2 n+1\}$, then $A$ is simple, and at this point $|A|=n+1$. We will prove that for any simple subset $A$, $|A| \leqslant n+1$. We use proof by contradiction. Assume $|A|>n+1$, then $A$ contains at least $n+2$ elements, let them be: $a_{1} < a_{2} < \cdots < a_{n+2}$. Consider the $2n+2$ numbers: $a_{1}, a_{2}, \cdots, a_{n+2}$ and $a_{2}-a_{1}, a_{3}-a_{1}, \cdots, a_{n+2}-a_{1}$. These are all positive integers not exceeding $2n+1$. By the pigeonhole principle, there must be two elements that are equal, and they can only be some $b_{i}$ and some $a_{j}-a_{1}$, thus $a_{1}+b_{i}=a_{j}$, so $A$ is not simple, a contradiction.

Method 2: Consider the $2n+2$ elements: $a_{2}<a_{3}<\cdots<a_{n+2}$ and $a_{2}-a_{1}<a_{3}-a_{1}<\cdots<a_{n+2}-a_{1}$, they are all positive integers not exceeding $2n+1$, so there must be two elements that are equal: let them be $a_{i}-a_{1}=a_{j}$. Therefore, $A$ is not simple, a contradiction.

Method 3: Consider the $2n+3$ elements: $a_{1}<a_{2}<a_{3}<\cdots<a_{n+2}$ and $a_{2}-a_{1}<a_{3}-a_{1}<\cdots<a_{n+2}-a_{1}$, they are all positive integers not exceeding $2n+1$. Note that $2n+3-(2n+1)=2$, so there must be two pairs of elements that are equal: $a_{i}-a_{1}=a_{j}, a_{s}-a_{1}=a_{t}$, and at least one of $a_{j}, a_{t}$ is not $a_{1}$, thus $A$ is not simple, a contradiction.

In conclusion, the maximum value of $|A|$ is $n+1$.