3 Let $n$ be a positive integer, and the set $M=\{1,2, \cdots, 2 n\}$. Find the smallest positive integer $k$, such that for any $k$-element subset of $M$, there must be 4 distinct elements whose sum equals $4 n+1$. (Supplied by Zhang Pengcheng)
Consider the $(n+2)$-element subset $P=\{n-1, n, n+1, \cdots, 2 n\}$ of $M$. The sum of any 4 different elements in $P$ is not less than $n-1+n+n+1+n+2=4 n+2$, so $k \geqslant n+3$.
Pair the elements of $M$ into $n$ pairs, $B_{i}=(i, 2 n+1-i), 1 \leqslant i \leqslant n$.
For any $(n+3)$-element subset $A$ of $M$, there must be three pairs $B_{i_{1}}, B_{i_{2}}, B_{i_{3}}$ all belonging to $A\left(i_{1}, i_{2}, i_{3}\right.$ are distinct).
Also pair the elements of $M$ into $n-1$ pairs, $C_{i}=(i, 2 n-i), 1 \leqslant i \leqslant n-1$.

For any $(n+3)$-element subset $A$ of $M$, there must be one pair $C_{i_{4}}$ belonging to $A$.
This pair $C_{i_{4}}$ must have no common element with at least one of the three pairs $B_{i_{1}}, B_{i_{2}}, B_{i_{3}}$, these 4 elements are distinct, and their sum is $2 n+1+2 n=4 n+1$.
Therefore, the smallest $k=n+3$.