21 *pure. Let $M=\{1,2, \cdots, 2 n+1\}, A$ be a subset of $M$, find the maximum value of $|A|$.

translated as:

21 *pure. Let $M=\{1,2, \cdots, 2 n+1\}, A$ be a subset of $M$, find the maximum value of $|A|$.
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 $A$, we have $|\Lambda| \leqslant n+1$. We use proof by contradiction. Assume $|A|>n+1$, then $A$ has at least $n+2$ elements, let them be: $a_{1}n$. positive even numbers. 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, which is a contradiction.
In conclusion, the maximum value of $|A|$ is $n+1$.