6. Given that $A$ and $B$ are two subsets of the set $\{1,2, \cdots, 100\}$, satisfying that $A$ and $B$ have the same number of elements, and $A \cap B$ is an empty set. If $n \in A$, then $2n+2 \in B$, the maximum number of elements in the set $A \cup B$ is ( ).
(A) 62
(B) 66
(C) 68
(D) 74
6. B.

First, we need to prove that $|A \cup B| \leqslant 66$, which only requires proving that $|A| \leqslant 33$. For this, we only need to prove: if $A$ is any 34-element subset of $\{1,2, \cdots, 49\}$, then there must exist $n \in A$ such that $2 n+2 \in A$. The proof is as follows:
Divide $\{1,2, \cdots, 49\}$ into the following 33 sets:
$\{1,4\},\{3,8\},\{5,12\}, \cdots,\{23,48\}$, a total of 12 sets;
$\{2,6\},\{10,22\},\{14,30\},\{18,38\}$, a total of 4 sets;
$\{25\},\{27\},\{29\}, \cdots,\{49\}$, a total of 13 sets;
$\{26\},\{34\},\{42\},\{46\}$, a total of 4 sets.
Since $A$ is a 34-element subset of $\{1,2, \cdots, 49\}$, by the pigeonhole principle, at least one of the 33 sets must contain two elements that both belong to $A$, i.e., there exists $n \in A$ such that $2 n+2 \in A$.

If we take $A=\{1,3,5, \cdots, 23,2,10,14,18,25,27,29, \cdots, 49,26,34,42,46\}, B=\{2 n+2 \mid n \in A\}$, then $A$ and $B$ satisfy the problem's conditions and $|A \cup B|=66$.