Three. (50 points) Let $M=\{1,2, \cdots, 65\}, A \subseteq M$ be a subset. If $|A|=33$, and there exist $x, y \in A, x<y$, $x \mid y$, then $A$ is called a "good set". Find the largest $a \in M$ such that any 33-element subset containing $a$ is a good set.
Three, Solution: Let $P=\{21+i \mid i=1,2, \cdots, 44\} \backslash\{2(21+i) \mid i=1,2, \cdots, 11\},|P|=33$. Clearly, for any $1 \leqslant i<j \leqslant 44$, there does not exist $n \geqslant 3$ such that $21+j=n(21+i)$ holds. Therefore, $P$ is a non-good set.

Thus, $a \leqslant 21$. We now prove that any 33-element subset $A$ containing 21 is definitely a good set. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{32}, 21\right\}$. If any of $1,3,7,42,63$ is an element of set $A$, it is clearly a good set.
Now consider the case where $1,3,7,42,63$ do not belong to set $A$. Construct the sets
$$
\begin{array}{l}
A_{1}=\{2,4,8,16,32,64\}, A_{2}=\{5,10,20,40\}, A_{3}=\{6,12,24,48\}, A_{4}=\{9,18,36\}, A_{5}= \\
\{11,22,44\}, A_{6}=\{13,26,52\}, A_{7}=\{14,28,56\}, A_{8}=\{15,30,60\}, A_{9}=\{17,34\}, \\
A_{10}=\{19,38\}, A_{11}=\{23,46\}, A_{12}=\{25,50\}, A_{13}=\{27,54\}, A_{14}=\{29,58\}, A_{15}=\{31,62\}, \\
A^{\prime}=\{33,35,37, \cdots, 61,65\} .
\end{array}
$$

From the above, it is clear that in each of $A_{1}, A_{2}, \cdots, A_{15}$, any two elements are in a multiple relationship. Consider the worst-case scenario, i.e., $A^{\prime} \subset A$, meaning all 16 elements of $A^{\prime}$ are chosen as elements of $A$. The remaining 16 elements of $A$ must be selected from the 15 sets $A_{1}, A_{2}, \cdots, A_{15}$. By the pigeonhole principle, at least one set will have two elements chosen, meaning that there are at least two elements in set $A$ that are in a multiple relationship.
In conclusion, any 33-element subset $A$ containing 21 is definitely a good set, i.e., the maximum value of $a$ is 21.