4. Let $n \in \mathbf{N}_{+}$, and the set $M=\{1,2, \cdots, 2 n\}$. There exists a positive integer $k$, such that for any $k$-element subset of $M$, there must be four distinct elements whose sum equals $4 n+1$, then $k_{\text {min }}=$ $\qquad$.
4. $n+3$.

Consider the $(n+2)$-element subset of $M$
$$
P=\{n-1, n, n+1, \cdots, 2 n\} \text {. }
$$

By the problem's condition, the sum of any four distinct 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:
$$
B_{i}=(i, 2 n+1-i), 1 \leqslant i \leqslant n \text {. }
$$

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:
$$
C_{i}=(i, 2 n-i), 1 \leqslant i \leqslant n-1 \text {. }
$$

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 elements with at least one of $B_{i_{1}}, B_{i_{2}}, B_{i_{3}}$, and these four elements are distinct, with their sum being $2 n+1+2 n=4 n+1$.
Thus, the smallest positive integer $k=n+3$.