9. (16 points) Let $n$ be a positive integer, and the set
$$
M=\{1,2, \cdots, 2 n\} \text {. }
$$

Find the smallest positive integer $k$, such that for any $k$-element subset of set $M$, there must be four distinct elements whose sum is $4 n+1$.
Consider the $(n+2)$-element subset of set $M$
$$
P=\{n-1, n, n+1, \cdots, 2 n\} \text {. }
$$

Since the sum of any four distinct elements in subset $P$ is not less than
$$
(n-1)+n+(n+1)+(n+2)=4 n+2,
$$

we know that $k \geqslant n+3$.
Pair the elements of set $M$ into $n$ pairs,
$$
B_{i}=(i, 2 n+1-i)(1 \leqslant i \leqslant n),
$$

For any $(n+3)$-element subset $A$ of set $M$, 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 set $M$ into $n-1$ pairs,
$$
C_{i}=(i, 2 n-i)(1 \leqslant i \leqslant n-1),
$$

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