Find the largest possible number of rooks that can be placed on a $3n \times 3n$ chessboard so that each rook is attacked by at most one rook.
To find the largest possible number of rooks that can be placed on a \(3n \times 3n\) chessboard so that each rook is attacked by at most one rook, we will proceed as follows:

1. **Base Case for \(n=1\)**:
    - For \(n=1\), the chessboard is \(3 \times 3\).
    - We can place 4 rooks in the squares \((1,1)\), \((1,2)\), \((2,3)\), and \((3,3)\).
    - This configuration ensures that each rook is attacked by at most one other rook.
    - To see why 4 is the maximum, note that if any row or column has 2 rooks, it forbids 5 other squares, leaving only 2 free squares.

2. **Inductive Step for \(n>1\)**:
    - Assume that for a \(3(n-1) \times 3(n-1)\) chessboard, the maximum number of rooks that can be placed such that each rook is attacked by at most one rook is \(4(n-1)\).
    - We need to show that for a \(3n \times 3n\) chessboard, the maximum number of rooks is \(4n\).

3. **Constructing the Configuration**:
    - Use the configuration described in the base case on each \(3 \times 3\) block along the diagonal of the \(3n \times 3n\) chessboard.
    - This gives us \(n\) blocks, each contributing 4 rooks, for a total of \(4n\) rooks.

4. **Upper Bound Argument**:
    - Consider a maximum configuration on a \(3n \times 3n\) chessboard.
    - Since \(M_n \geq 4n > 3n\), there must be at least one row and one column with 2 rooks.
    - By swapping rows and columns, we can arrange the board such that the first row and first column each contain 2 rooks.
    - Place these rooks in positions \((1,1)\), \((1,2)\), \((2,3)\), and \((3,3)\).
    - No other rooks can be in the first three rows or columns.
    - The remaining \(3(n-1) \times 3(n-1)\) sub-board must contain \(M_n - 4\) rooks.
    - By the inductive hypothesis, \(M_n - 4 \leq M_{n-1}\), so \(M_n \leq M_{n-1} + 4\).

5. **Conclusion**:
    - Combining the lower bound \(M_n \geq 4n\) and the upper bound \(M_n \leq M_{n-1} + 4\), we get \(M_n = 4n\).

\(\blacksquare\)

The final answer is \( \boxed{ 4n } \).