For positive integers $n$, let $M_n$ be the $2n+1$ by $2n+1$ skew-symmetric matrix for which each entry in the first $n$ subdiagonals below the main diagonal is 1 and each of the remaining entries below the main diagonal is -1. Find, with proof, the rank of $M_n$. (According to one definition, the rank of a matrix is the largest $k$ such that there is a $k \times k$ submatrix with nonzero determinant.)

One may note that
\begin{align*}
M_1 &= \left( \begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0
\end{array}\right) \\
M_2 &= \left( \begin{array}{ccccc} 0 & -1 & -1 & 1
& 1 \\ 1 & 0 & -1 & -1 & 1 \\ 1 & 1 & 0 & -1 & -1 \\ -1 & 1 & 1 & 0 &
-1 \\ -1 & -1 & 1 & 1 & 0 \end{array} \right).
\end{align*}
1. **Matrix Definition and Properties**:
   - We are given a \( (2n+1) \times (2n+1) \) skew-symmetric matrix \( M_n \).
   - Each entry in the first \( n \) subdiagonals below the main diagonal is 1.
   - Each of the remaining entries below the main diagonal is -1.
   - Since \( M_n \) is skew-symmetric, \( M_n^T = -M_n \).

2. **Rank Upper Bound**:
   - The sum of all rows of \( M_n \) is zero. This implies that the rows are linearly dependent.
   - Therefore, the rank of \( M_n \) is at most \( 2n \).

3. **Rank Lower Bound**:
   - To show that the rank is exactly \( 2n \), we need to show that the kernel of \( M_n \) is one-dimensional.
   - Suppose \( \mathbf{x} = (x_1, x_2, \ldots, x_{2n+1}) \) is in the kernel of \( M_n \), i.e., \( M_n \mathbf{x} = \mathbf{0} \).

4. **Kernel Analysis**:
   - By the definition of \( M_n \), for each \( k \), we have:
     \[
     x_{k+1} + x_{k+2} + \cdots + x_{k+n} = x_{k+n+1} + x_{k+n+2} + \cdots + x_{k+2n}
     \]
     where indices are considered modulo \( 2n+1 \).

5. **Equating Sums**:
   - This implies:
     \[
     x_{k+1} + x_{k+2} + \cdots + x_{k+n} = x_{k+n+1} + x_{k+n+2} + \cdots + x_{k+2n}
     \]
     \[
     x_{k+1} + x_{k+2} + \cdots + x_{k+n} = x_{k+2n+1} + x_{k+2n+2} + \cdots + x_{k+3n} = x_k + x_{k+1} + \cdots + x_{k+n-1}
     \]
     \[
     \Rightarrow x_k = x_{k+n}
     \]

6. **Conclusion**:
   - From \( x_k = x_{k+n} \), we see that \( x_{k+1} = x_{k+n+1} = x_{k+2n+1} = x_k \).
   - Therefore, all \( x_i \) are equal, implying that the kernel is one-dimensional.

7. **Rank Determination**:
   - Since the kernel is one-dimensional, the rank of \( M_n \) is \( 2n \).

\(\blacksquare\)

The final answer is \( \boxed{ 2n } \)