We are placing rooks on a $n \cdot  n$ chess table that providing this condition:
Every two rooks will threaten an empty square at least.
What is the most number of rooks?
1. **Claim**: For an odd \( n \), the maximum number of rooks that can be placed on an \( n \times n \) chessboard such that every two rooks will threaten an empty square is \(\frac{3n-1}{2}\). For an even \( n \), the maximum number is \(\frac{3n-2}{2}\).

2. **Construction for odd \( n \)**: 
   - Place \( n \) rooks on the main diagonal.
   - Place \(\frac{n-1}{2}\) rooks on the first half of the other diagonal.

3. **Proof by contradiction**:
   - Suppose we can place more than \(\frac{3n-1}{2}\) rooks on the board.
   - Since no row or column can have more than 2 rooks, there must be at least \(\frac{n+1}{2}\) rows with 2 rooks each.

4. **Lemma**: There cannot be more than \(\frac{n-1}{2}\) rows with more than 1 rook.
   - **Proof**: Suppose the rooks in those rows are ordered by the distance \( d_i \) between two of them. Note these distances all, if ordered in a list, show us that \( d_{i-1} - d_i \geq 2 \). This is because if two rooks are placed closer than this, they would threaten a common square, violating the given condition. Therefore, the number of rows with more than 1 rook is limited to \(\frac{n-1}{2}\).

5. **Conclusion**: The assumption that we can place more than \(\frac{3n-1}{2}\) rooks leads to a contradiction. Hence, the maximum number of rooks that can be placed on an \( n \times n \) chessboard for odd \( n \) is \(\frac{3n-1}{2}\).

The final answer is \( \boxed{ \frac{3n-1}{2} } \) for odd \( n \) and \(\frac{3n-2}{2}\) for even \( n \).