What is the least posssible number of cells that can be marked on an $n \times n$ board such that for each $m >\frac{ n}{2}$ both diagonals of any $m \times m$  sub-board contain a marked cell?
1. **Understanding the Problem:**
   We need to find the minimum number of cells that must be marked on an \( n \times n \) board such that for any \( m \times m \) sub-board with \( m > \frac{n}{2} \), both diagonals of the sub-board contain at least one marked cell.

2. **Analyzing the Example:**
   - For odd \( n = 2k + 1 \):
     - Mark the central row, which is the \((k+1)\)-th row.
   - For even \( n = 2k \):
     - Mark the \((k+1)\)-th row.

3. **Proof for Odd \( n = 2k + 1 \):**
   - Consider any \( m \times m \) sub-board with \( m > \frac{n}{2} \). Since \( n = 2k + 1 \), we have \( \frac{n}{2} = k + \frac{1}{2} \), so \( m > k + \frac{1}{2} \) implies \( m \geq k + 1 \).
   - The central row (the \((k+1)\)-th row) intersects every \( m \times m \) sub-board with \( m \geq k + 1 \) and thus ensures that both diagonals of any such sub-board contain a marked cell.

4. **Proof for Even \( n = 2k \):**
   - Similarly, for \( n = 2k \), \( \frac{n}{2} = k \), so \( m > k \) implies \( m \geq k + 1 \).
   - Marking the \((k+1)\)-th row ensures that every \( m \times m \) sub-board with \( m \geq k + 1 \) will have a marked cell on both diagonals.

5. **Proof that Fewer than \( n \) Marked Cells are Insufficient:**
   - Suppose we mark fewer than \( n \) cells. Consider \( n \) disjoint diagonals of the \( n \times n \) board. Each diagonal can be part of an \( m \times m \) sub-board with \( m > \frac{n}{2} \).
   - If we have fewer than \( n \) marked cells, at least one of these \( n \) diagonals will not contain a marked cell, violating the condition that both diagonals of any \( m \times m \) sub-board must contain a marked cell.

Thus, the least possible number of cells that can be marked is \( n \).

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