On a $300 \times 300$ board, several rooks are placed that beat the entire board. Within this case, each rook beats no more than one other rook.  At what least $k$, it is possible to state that there is at least one rook in each $k\times k$ square ?
1. **Generalization and Initial Setup**:
   - We are given a $300 \times 300$ board and need to determine the smallest $k$ such that every $k \times k$ square contains at least one rook.
   - The problem generalizes to a $3n \times 3n$ board, where $n \in \mathbb{Z}_{>0}$.
   - The answer is $\boxed{2n+1}$, and for $n=100$, the answer is $201$.

2. **Construction for $k=2n$**:
   - We need to show that $k=2n$ does not work by providing a construction where a $2n \times 2n$ square does not contain any rooks.
   - For $n=3$ and $n=4$, we can construct examples where rooks are placed such that no $2n \times 2n$ square contains a rook.

3. **Example for $n=3$**:
   - Consider a $9 \times 9$ board (since $3n = 9$).
   - Place rooks in positions such that no $6 \times 6$ square contains a rook.
   - The construction is as follows:
     - Place rooks at $(1,8)$, $(2,7)$, $(3,6)$, $(4,5)$, $(5,4)$, $(6,3)$, $(7,2)$, $(8,1)$, and $(9,9)$.
     - This ensures that no $6 \times 6$ square contains a rook.

4. **Example for $n=4**:
   - Consider a $12 \times 12$ board (since $3n = 12$).
   - Place rooks in positions such that no $8 \times 8$ square contains a rook.
   - The construction is as follows:
     - Place rooks at $(1,11)$, $(2,10)$, $(3,9)$, $(4,8)$, $(5,7)$, $(6,6)$, $(7,5)$, $(8,4)$, $(9,3)$, $(10,2)$, $(11,1)$, and $(12,12)$.
     - This ensures that no $8 \times 8$ square contains a rook.

5. **Proof for $k=2n+1$**:
   - Assume for contradiction that there exists a $(2n+1) \times (2n+1)$ square that does not contain any rooks.
   - Without loss of generality, assume this square is in the bottom left corner of the board.
   - Let $A$ be the $(2n+1) \times (n-1)$ rectangle on top of the $(2n+1) \times (2n+1)$ square.
   - Let $B$ be the $(n-1) \times (2n+1)$ rectangle to the right of the $(2n+1) \times (2n+1)$ square.
   - By assumption, the $(2n+1) \times (2n+1)$ square does not contain any rooks.
   - The rectangle $A$ can contain at most $2n-2$ rooks (since each row can contain at most one rook).
   - Therefore, there must be at least one row in $A$ that does not contain any rooks.
   - Similarly, the rectangle $B$ can contain at most $2n-2$ rooks, so there must be at least one column in $B$ that does not contain any rooks.
   - The intersection of this row and column is a unit square that is not attacked by any rook, which contradicts the assumption that the rooks beat the entire board.
   - Hence, every $(2n+1) \times (2n+1)$ square must contain at least one rook.