How many rooks can be placed in an $n\times n$ chessboard such that each rook is threatened by at most $2k$ rooks?
(15 points)
[i]Proposed by Mostafa Einollah zadeh[/i]
1. **Define the problem and variables:**
   We need to determine the maximum number of rooks that can be placed on an \( n \times n \) chessboard such that each rook is threatened by at most \( 2k \) other rooks. Let \( x \) be the total number of rooks placed on the board. For each rook, let its score be the number of rooks it threatens. 

2. **Introduce row and column variables:**
   Let \( r_i \) be the number of rooks in the \( i \)-th row and \( c_i \) be the number of rooks in the \( i \)-th column. The total sum of the scores of all the rooks, denoted by \( s \), can be expressed as:
   \[
   s = \sum_{i=1}^{n} r_i(r_i - 1) + \sum_{i=1}^{n} c_i(c_i - 1)
   \]

3. **Double counting and convexity:**
   By double counting, each pair of rooks that threaten each other is counted twice, once in the row sum and once in the column sum. Therefore, the total sum of the scores \( s \) is at most \( 2kx \):
   \[
   s \leq 2kx
   \]

4. **Estimate the sum of squares:**
   Using the Cauchy-Schwarz inequality, we can estimate the sum of the squares of the row and column counts:
   \[
   \sum_{i=1}^{n} r_i^2 \geq \frac{(\sum_{i=1}^{n} r_i)^2}{n} = \frac{x^2}{n}
   \]
   Similarly,
   \[
   \sum_{i=1}^{n} c_i^2 \geq \frac{x^2}{n}
   \]

5. **Combine the inequalities:**
   Since \( r_i(r_i - 1) = r_i^2 - r_i \) and \( c_i(c_i - 1) = c_i^2 - c_i \), we have:
   \[
   \sum_{i=1}^{n} r_i(r_i - 1) + \sum_{i=1}^{n} c_i(c_i - 1) = \sum_{i=1}^{n} (r_i^2 - r_i) + \sum_{i=1}^{n} (c_i^2 - c_i)
   \]
   \[
   = \sum_{i=1}^{n} r_i^2 + \sum_{i=1}^{n} c_i^2 - x
   \]
   Using the previous estimates:
   \[
   \sum_{i=1}^{n} r_i^2 + \sum_{i=1}^{n} c_i^2 \geq 2 \cdot \frac{x^2}{n}
   \]
   Therefore,
   \[
   s \geq 2 \cdot \frac{x^2}{n} - x
   \]

6. **Set up the inequality:**
   Combining the upper and lower bounds for \( s \):
   \[
   2 \cdot \frac{x^2}{n} - x \leq 2kx
   \]
   Simplifying, we get:
   \[
   2 \cdot \frac{x^2}{n} - x \leq 2kx
   \]
   \[
   2 \cdot \frac{x^2}{n} \leq (2k + 1)x
   \]
   \[
   \frac{x^2}{n} \leq (k + \frac{1}{2})x
   \]
   \[
   x \leq n(k + 1)
   \]

7. **Achieving equality:**
   Equality is achieved when there are exactly \( k+1 \) rooks in each row and column. This can be arranged by placing rooks in the \( i, i+1, \cdots, i+k \)-th squares of the \( i \)-th row for all \( i \), with indices taken modulo \( n \).

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