6. Given $S=\{1,2, \cdots, 216\}, A \subseteq S$. If the sum of any two elements in set $A$ cannot be divisible by 6, then the maximum number of elements in set $A$ is ( ).
(A) 36
(B) 52
(C) 74
(D) 90
$$
\begin{array}{c}
\text { 6.C. } \\
\text { Let } S_{k}=\{x \in S \mid x=6 n+k, n \in \mathbb{N}(k= \\
0,1, \cdots, 5) \text {, and } S=\bigcup_{k=0}^{5} S_{k} .
\end{array}
$$

It is known that $\operatorname{card}\left(S_{k}\right)=36$. Then, in set $A$, it cannot simultaneously contain elements from $S_{1}$ and $S_{5}$, or $S_{2}$ and $S_{4}$, nor can it have two elements from $S_{6}$ or two elements from $S_{3}$. To maximize the number of elements in $A$, one can select all elements from $S_{1}$ and $S_{2}$, and one element each from $S_{0}$ and $S_{3}$. Therefore, the maximum number of elements is $36+36+1+1=74$.