Let $n$ and $k$ be positive integers satisfying $k\leq2n^2$. Lee and Sunny play a game with a $2n\times2n$ grid paper. First, Lee writes a non-negative real number no greater than $1$ in each of the cells, so that the sum of all numbers on the paper is $k$. Then, Sunny divides the paper into few pieces such that each piece is constructed by several complete and connected cells, and the sum of all numbers on each piece is at most $1$. There are no restrictions on the shape of each piece. (Cells are connected if they share a common edge.)
Let $M$ be the number of pieces. Lee wants to maximize $M$, while Sunny wants to minimize $M$. Find the value of $M$ when Lee and Sunny both play optimally. 
1. **Initial Setup and Objective**:
   - We are given a \(2n \times 2n\) grid.
   - Lee writes non-negative real numbers in each cell such that the sum of all numbers is \(k\) and each number is at most 1.
   - Sunny divides the grid into pieces such that the sum of numbers in each piece is at most 1.
   - Lee aims to maximize the number of pieces \(M\), while Sunny aims to minimize \(M\).

2. **Optimal Strategy for Lee**:
   - Lee can place the number \(\frac{k}{2k-1}\) in \(2k-1\) cells and 0 in the remaining cells.
   - This ensures that the sum of numbers in the grid is \(k\) and each number is at most 1.
   - Any piece Sunny creates can contain at most one non-zero number, leading to at least \(2k-1\) pieces.

3. **Optimal Strategy for Sunny**:
   - Sunny needs to ensure that the number of pieces \(M\) is minimized.
   - We need to show that \(2k-1\) pieces are sufficient for Sunny to achieve this.

4. **Proof of Sufficiency**:
   - We assume that if the sum of numbers in a piece is \(\leq 1\), it contains at most one non-zero number.
   - Let \(f(i)\) be the number of non-zero cells in the \(i\)-th row and \(s(i)\) be the sum of numbers in the \(i\)-th row.
   - We claim that for any \(1 \leq i \leq 2n-1\), \(s(i) + s(i+1) > \frac{f(i) + f(i+1)}{2}\).

5. **Detailed Proof**:
   - Fix any \(i \in \{1, 2, \ldots, 2n-1\}\).
   - Let \(a_1, a_2, \ldots, a_k\) be the non-zero numbers in the \(i\)-th row and \(b_1, b_2, \ldots, b_l\) be the non-zero numbers in the \((i+1)\)-th row.
   - Consider the connected pieces:
     - \(a_j\) and \(a_{j+1}\) for \(1 \leq j \leq k-1\).
     - \(b_j\) and \(b_{j+1}\) for \(1 \leq j \leq l-1\).
     - \(a_1\) and \(b_1\).
     - \(a_k\) and \(b_l\).

6. **Summing the Pieces**:
   \[
   2 \left( s(i) + s(i+1) \right) = 2 \left( \sum_{j=1}^{k} a_j + \sum_{j=1}^{l} b_j \right)
   \]
   \[
   = \sum_{j=1}^{k-1} (a_j + a_{j+1}) + \sum_{j=1}^{l-1} (b_j + b_{j+1}) + (a_1 + b_1) + (a_k + b_l)
   \]
   \[
   > \sum_{j=1}^{k-1} 1 + \sum_{j=1}^{l-1} 1 + 1 + 1 = (k-1) + (l-1) + 1 + 1 = k + l = f(i) + f(i+1)
   \]

7. **Conclusion**:
   - Summing over all rows:
   \[
   k = \left( s(1) + s(2) \right) + \left( s(3) + s(4) \right) + \cdots + \left( s(2n-1) + s(2n) \right)
   \]
   \[
   > \left( \frac{f(1) + f(2)}{2} \right) + \left( \frac{f(3) + f(4)}{2} \right) + \cdots + \left( \frac{f(2n-1) + f(2n)}{2} \right)
   \]
   \[
   = \frac{f(1) + f(2) + \cdots + f(2n)}{2} = \frac{\# \text{ non-zero numbers in the table}}{2}
   \]
   - Therefore, the number of non-zero numbers in the table is at most \(2k-1\).
   - Thus, \(2k-1\) pieces suffice.

This completes the proof of the problem. \(\blacksquare\)

The final answer is \( \boxed{ 2k-1 } \)