Example $1 \quad X=\{1,2,3, \cdots, 2 n+1\} . A$ is a subset of $X$, with the property: the sum of any two numbers in $A$ is not in $A$, find $\max |A|$.
Let's first consider a specific $A$. In the problem, $A$ is a set of numbers with a certain property, but the sum of any two numbers in $A$ does not have this property.

When giving examples, start with the familiar. The most common property of integers is parity, and the sum of any two odd numbers is not odd. Therefore,
$$
A=\{1,3,5, \cdots, 2 n+1\}
$$

is an example that meets the requirements. At this time,
$$
|A|=n+1.
$$
$A$ clearly cannot include any other numbers; adding any even number would no longer meet the requirements. And removing some odd numbers seems to prevent adding more even numbers (this will be strictly proven below). Therefore, we boldly guess
$$
\max |A|=n+1 \text{. }
$$

In Example 1, parity is very helpful. For a general $A$ that meets the requirements, assume there are $k(k \leqslant n+1)$ odd numbers in it:
$$
a_{1}>a_{2}>\cdots>a_{k}.
$$

Now let's see how many even numbers can be in $A$. Clearly, the even numbers
$$
a_{1}-a_{2}<a_{1}-a_{3}<\cdots<a_{1}-a_{k}
$$

cannot be in $A$ (because $a_{1}-a_{1}$ and $a_{i}$ sum to $a_{1}$, which is in $A$), so $A$ can have at most
$$
n-(k-1)=n+1-k
$$

even numbers. Therefore,
$$
|A| \leqslant k+(n+1-k)=n+1.
$$