6. Given that $A$ and $B$ are two subsets of the set $\{1,2,3, \cdots, 100\}$, satisfying: $A$ and $B$ have the same number of elements, and $A \cap B$ is an empty set. If $n \in A$ then $2 n+2 \in B$, the maximum number of elements in the set $A \cup B$ is
A. 62
B. 66
C. 68
D. 74
Since $2(49+1)=100$, the largest element in set $A$ does not exceed 49.
Take $A=\{49,48, \cdots, 24,10,9, \cdots, 5,1\}, B=\{100,98, \cdots, 50,22,20, \cdots, 12,4\}$, at this point $|A|=|B|=33,|A \bigcup B|=66$.
On the other hand, group the elements in $\{1,2,3, \cdots, 100\}$ that are less than or equal to 49 as follows: $(49),(48,23)$, $(47),(46,22), \cdots,(25),(24,11),(10,4),(9),(8,3),(7),(6,2),(5),(4,1)$, a total of 33 groups. When $A$ has more than 34 elements, there must be two elements from the same group, at which point $A \bigcap B \neq \varnothing$, leading to a contradiction. Therefore, the number of elements in set $A \cup B$ is at most 66. So the answer is $B$.