Four, in an election, there are 12 candidates, and each member of the electoral committee casts 6 votes. It is known that any two members' votes have at most 2 candidates in common. Find the maximum number of members in the committee.
(Proposed by the Problem Committee)
The maximum number of committee members is 4.
Let the number of committee members be $k$, and the candidates be represented by $1,2, \cdots, 12$. Each person's vote is a set $A_{i}(1 \leqslant i \leqslant k)$, and the number of votes each candidate receives is $m_{i}(1 \leqslant i \leqslant 12)$. Then,
$$
\sum_{i=1}^{12} m_{i}=\sum_{i=1}^{k} 6=6 k .
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
2 \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right|=2 \sum_{i=1}^{12} \mathrm{C}_{m_{i}}^{2}=\sum_{i=1}^{12} m_{i}^{2}-\sum_{i=1}^{12} m_{i} \\
\geqslant \frac{\left(\sum_{i=1}^{12} m_{i}\right)^{2}}{\sum_{i=1}^{12} 1^{2}}-\sum_{i=1}^{12} m_{i}=\frac{(6 k)^{2}}{12}-6 k \\
=3 k^{2}-6 k .
\end{array}
$$

Notice that,
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right| \leqslant \sum_{1 \leqslant i<j \leqslant k} 2=2 \mathrm{C}_{k}^{2}=k^{2}-k \\
\Rightarrow & 3 k^{2}-6 k \leqslant 2\left(k^{2}-k\right) \Rightarrow k \leqslant 4 .
\end{aligned}
$$

When $k=4$, a construction that satisfies the conditions is
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5,6\}, \\
A_{2}=\{1,2,7,8,9,10\}, \\
A_{3}=\{3,4,7,8,11,12\}, \\
A_{4}=\{5,6,9,10,11,12\} .
\end{array}
$$