2.99 A set of numbers is called "simple" if it does not contain elements $x, y, z$ such that $x+y=z$. Given the set $\{1,2, \cdots, 2 n+1\}$. Find the maximum number of elements a simple subset can contain.
[Solution] Consider the two subsets of $\{1,2, \cdots, 2 n+1\}$:
$$
\begin{array}{l}
A=\{1,3,5, \cdots, 2 n+1\}, \\
B=\{2,4,6, \cdots, 2 n\} .
\end{array}
$$

Since the sum of any two numbers in set $A$ is even, there are no elements $x, y, z$ such that $x+y=z$, so set $A$ is a simple subset. The number of elements in $A$ is $n+1$.
Next, we prove that $n+1$ is the maximum number of elements in such a subset.
Suppose $\left\{a_{0}, a_{1}, \cdots, a_{n+1}\right\}$ is a subset of $\{1,2, \cdots, 2 n+1\}$ with $n+2$ elements. We will prove that it is not a simple subset, i.e., there exist three elements $x, y, z$ such that $x+y=z$.
Let $a_{0}<a_{1}<a_{2}<\cdots<a_{n+1} \leqslant 2 n+1$,
and define $b_{i}=a_{i}-a_{0}, i=1,2, \cdots, n+1$, then
$$
b_{1}<b_{2}<b_{3}<\cdots<b_{n+1} \leqslant 2 n+1 .
$$

For the $2 n+3$ positive integers $a_{0}, a_{1}, \cdots, a_{n+1}$ and $b_{1}, b_{2}, \cdots, b_{n+1}$, which are all no greater than $2 n+1$, there must be two numbers that are equal. Let $a_{i}=b_{j}$, i.e.,
$$
a_{i}=b_{j}=a_{j}-a_{0},
$$

thus
$$
a_{i}+a_{0}=a_{j} .
$$

Let $a_{i}=x, a_{0}=y, a_{j}=z$, then $x+y=z$, which means the set $\left\{a_{0}, a_{1}, \cdots, a_{n+1}\right\}$ with $n+2$ elements is not a simple subset.
Therefore, $n+1$ is the maximum number that meets the requirements of the problem.