Consider the $(n+2)$-element subset $P=\{n-1, n, n+1, \cdots, 2 n\} \cdot P$. The sum of any 4 different elements in $P$ is not less than $n-1+n+n+1+n+2=4 n+1$, 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 $B_{i 1}, B_{i 2}, B_{i 3}$ that all belong to $A\left(i_{1}, i_{2}, i_{3}\right.$ are all different).
Also, pair the elements of $M$ into $n-1$ pairs, $C_{i}=(i, 2 n-i), 1 \leqslant i \leqslant n-1$. For any $\cdots n+3$-element subset $A$ of $M$, there must be one pair $C_{i 4}$ that belongs 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 all different, and their sum is $2 n+1+2 n=4 n+1$.
Therefore, the smallest positive integer $k=n+3$.
Example 4 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 $4 n+1$.
(2005 Southeastern Mathematical Olympiad Problem)