Given an odd $n\in\mathbb N$. In an $n\times n$ chessboard, you may place many $2\times2$ squares. How many grids, at most, are covered by exactly one square?
1. **Initial Assumption and Contradiction:**
   - Assume there exists a column with \( n \) grids, such that each grid is covered by exactly one \( 2 \times 2 \) square.
   - Since \( n \) is odd, covering \( n \) grids in pairs (each pair being part of a \( 2 \times 2 \) square) is impossible because \( n \) cannot be evenly divided into pairs. This leads to a contradiction.

2. **Upper Bound Calculation:**
   - Since each column can contain at most \( n-1 \) grids covered by exactly one \( 2 \times 2 \) square, the total number of such grids on the board is at most \( n(n-1) \).

3. **Constructive Proof for Achievability:**
   - For \( n = 3 \), consider a \( 3 \times 3 \) chessboard. It is easy to see that we can place \( 2 \times 2 \) squares such that some grids are covered exactly once.
   - For a general odd \( n \), consider an L-shaped block composed of two rectangles of length \( n+2 \) and width \( 2 \). Combining this L-shaped block with an \( n \times n \) block results in an \( (n+2) \times (n+2) \) block.
   - Cover the grids in the block by successively placing the \( 2 \times 2 \) squares from left to right and top to bottom, ensuring that only the two squares at the turn overlap at exactly one point.
   - This method covers \( 8 \times \frac{n+1}{2} - 2 \) grids, which simplifies to \( 4n + 2 \) grids covered exactly once.

4. **Inductive Step:**
   - By induction, if the desired configuration can be achieved for \( n \), it can also be achieved for \( n+2 \).

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