Let $n$ be positive integer, set $M = \{ 1, 2, \ldots, 2n \}$. Find the minimum positive integer $k$ such that for any subset $A$ (with $k$ elements) of set $M$, there exist four pairwise distinct elements in $A$ whose sum is $4n + 1$.
1. **Define the problem and set up the notation:**
   Let \( n \) be a positive integer, and consider the set \( M = \{ 1, 2, \ldots, 2n \} \). We need to find the minimum positive integer \( k \) such that for any subset \( A \) of \( M \) with \( k \) elements, there exist four pairwise distinct elements in \( A \) whose sum is \( 4n + 1 \).

2. **Initial observation:**
   Consider the subset \( \{n-1, n, n+1, n+2, \ldots, 2n\} \). This subset has \( n+2 \) elements. The minimum sum of any four distinct elements in this subset is:
   \[
   (n-1) + n + (n+1) + (n+2) = 4n + 2
   \]
   Since \( 4n + 2 \) is greater than \( 4n + 1 \), this subset does not contain four elements that sum to \( 4n + 1 \). Therefore, \( k \) must be at least \( n+3 \).

3. **Prove that \( k = n+3 \) is sufficient:**
   We need to show that any subset \( A \) of \( M \) with \( n+3 \) elements contains four pairwise distinct elements whose sum is \( 4n + 1 \).

4. **Using the Pigeonhole Principle:**
   - Consider a subset \( A \) of \( M \) with \( n+3 \) elements.
   - We can form pairs of distinct elements from \( A \). There are \( \binom{n+3}{2} = \frac{(n+3)(n+2)}{2} \) pairs.
   - The possible sums of these pairs range from \( 3 \) (smallest sum: \( 1+2 \)) to \( 4n-1 \) (largest sum: \( (2n-1) + 2n \)).

5. **Finding pairs that sum to \( 2n \):**
   - Since \( A \) has \( n+3 \) elements, there are \( \binom{n+3}{2} \) pairs, which is more than \( 2n-1 \) possible sums.
   - By the Pigeonhole Principle, there must be at least one pair of elements in \( A \) that sum to \( 2n \).

6. **Removing the pair and finding the next pair:**
   - Remove the pair of elements that sum to \( 2n \) from \( A \). We are left with \( n+1 \) elements.
   - Now, consider the remaining \( n+1 \) elements. We can form pairs from these elements, and by the Pigeonhole Principle, there must be at least one pair of elements that sum to \( 2n+1 \).

7. **Conclusion:**
   - We have found two pairs of elements: one pair that sums to \( 2n \) and another pair that sums to \( 2n+1 \).
   - The sum of these four elements is:
     \[
     2n + (2n+1) = 4n + 1
     \]
   - Therefore, any subset \( A \) of \( M \) with \( n+3 \) elements contains four pairwise distinct elements whose sum is \( 4n + 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ k = n+3 } \)