Three, let $n(n \geqslant 3)$ 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 $4 n+1$.
Three: Consider the $(n+2)$-element subset of $M$
$$
P=\{n-1, n, n+1, \cdots, 2 n\} \text {. }
$$
The sum of any 4 different elements in $P$ is not less than
$$
n-1+n+n+1+n+2=4 n+2 \text {, }
$$

Therefore, $k \geqslant n+3$.
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 $A\left(i_{1} 、 i_{2} 、 i_{3}\right.$ 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 $(n+3)$-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 all different, and their sum is
$$
2 n+1+2 n=4 n+1 \text {. }
$$

Therefore, the smallest positive integer $k=n+3$.