Question 13: Given that $A$ and $B$ are two subsets of $\{1,2, \ldots, 100\}$, satisfying: $|A|=|B|, A \cap B=\emptyset$, and for any $x \in A, 2 x+2 \in B$. Try to find the maximum value of $|A \cup B|$.

---

The translation maintains the original format and line breaks as requested.
Question 13, Solution: First, prove: $|A \cup B| \leq 66$, which is equivalent to proving: $|A| \leq 33$. For this, it suffices to prove: if $A$ is any 34-element subset of $\{1,2, \ldots, 49\}$, then there must exist $n \in A$ such that $2n+2 \in A$.

In fact, partition $\{1,2, \ldots, 49\}$ into the following 33 subsets:
$$
\begin{array}{l}
\{1,4\},\{3,8\},\{5,12\}, \ldots,\{23,48\} ; \\
\{2,6\},\{10,22\},\{14,30\},\{18,38\} ; \\
\{25\},\{27\}, \ldots\{49\} ; \\
\{26\},\{34\},\{42\},\{46\}
\end{array}
$$

According to the pigeonhole principle, when selecting any 34 elements from $\{1,2, \ldots, 49\}$, there must be two numbers that fall into the same subset. On the other hand, take
$$
\begin{array}{l}
A=\{1,3, \ldots, 23\} \cup\{2,10,14,18\} \cup\{25,27, \ldots, 49\} \cup\{26,34,42,46\}, \\
B=\{2x \mid x \in A\} \\
\text { satisfying the conditions, and }|A \cup B|=66.
\end{array}
$$

Thus, the conditions are satisfied, and $|A \cup B|=66$.
In summary, the maximum value of $|A \cup B|$ is 66.