Let $n$ be a natural number. At first the cells of a table $2n$ x $2n$ are colored in white. Two players $A$ and $B$ play the following game. First is $A$ who has to color $m$ arbitrary cells in red and after that $B$ chooses $n$ rows and $n$ columns and color their cells in black. Player $A$ wins, if there is at least one red cell on the board. Find the least value of $m$ for which $A$ wins no matter how $B$ plays.
1. **Understanding the Problem:**
   - We have a \(2n \times 2n\) table initially colored white.
   - Player \(A\) colors \(m\) arbitrary cells red.
   - Player \(B\) then chooses \(n\) rows and \(n\) columns and colors their cells black.
   - Player \(A\) wins if there is at least one red cell remaining on the board.

2. **Objective:**
   - Determine the minimum value of \(m\) such that Player \(A\) always wins, regardless of Player \(B\)'s choices.

3. **Strategy and Analysis:**
   - Player \(B\) will try to cover as many red cells as possible by choosing \(n\) rows and \(n\) columns.
   - The intersection of these \(n\) rows and \(n\) columns forms an \(n \times n\) subgrid.
   - The total number of cells covered by \(B\) is \(2n \times n - n^2 = n(2n - n) = n^2\).

4. **Counting the Cells:**
   - The \(2n \times 2n\) table has \(4n^2\) cells.
   - Player \(B\) covers \(n \times 2n + n \times 2n - n^2 = 2n^2\) cells (since the \(n^2\) cells in the intersection are counted twice).

5. **Ensuring a Red Cell Remains:**
   - To ensure at least one red cell remains, Player \(A\) must place enough red cells such that not all can be covered by \(B\).
   - If \(B\) can cover \(2n^2\) cells, then \(A\) must place more than \(2n^2\) red cells to guarantee at least one remains uncovered.

6. **Calculating the Minimum \(m\):**
   - To ensure at least one red cell remains, \(A\) must place \(m\) red cells such that \(m > 2n^2\).
   - The smallest integer greater than \(2n^2\) is \(2n^2 + 1\).

7. **Verification:**
   - If \(m = 2n^2 + 1\), then even if \(B\) covers \(2n^2\) cells, there will be at least one red cell left uncovered.

Thus, the least value of \(m\) for which \(A\) wins no matter how \(B\) plays is \(2n^2 + 1\).

The final answer is \(\boxed{2n^2 + 1}\).