Let $n$ a positive integer. In a $2n\times 2n$ board,  $1\times n$ and $n\times 1$ pieces are arranged without overlap. 
Call an arrangement [b]maximal[/b] if it is impossible to put a new piece in the board without overlapping the previous ones.
Find the least $k$ such that there is a [b]maximal[/b] arrangement that uses $k$ pieces.
To solve this problem, we need to find the least number \( k \) such that there is a maximal arrangement of \( 1 \times n \) and \( n \times 1 \) pieces on a \( 2n \times 2n \) board. A maximal arrangement means that no additional piece can be placed without overlapping an existing piece.

1. **Construction for \( n \geq 3 \)**:
   - Number the rows and columns from 1 to \( 2n \), left to right and top to bottom.
   - Place a vertical tile with the top end at row 2, column 1.
   - Place \( 2n \) horizontal tiles in each row with the left end in column 2.
   - Slide the horizontal tiles in row 2 and row \( n+1 \) all the way to the right.

   This construction uses \( 2n + 1 \) tiles.

2. **Case Analysis**:
   - **Case 1: No tiles of one orientation**:
     - Suppose there are no horizontal tiles. Then every column must have a vertical tile.
     - Any column whose vertical tile touches an edge of the grid must have a second vertical tile.
     - At least two vertical tiles must touch each edge, so we have at least \( 2n + 2 \) tiles in this case.
     - The case where there are no vertical tiles is identical.

   - **Case 2: Tiles of both orientations**:
     - Suppose there are at most \( n \) vertical tiles.
     - Suppose for \( 1 < k \leq n \), we have a vertical tile in column \( k \).
     - Slide this tile left one space. The tiling is maximal, so it must now intersect another tile.
     - It can't be a horizontal tile since there is not enough space, so it must be vertical, and we conclude there is a vertical tile in column \( k-1 \).
     - Similarly, for any vertical tile in column \( n+1 \leq k < 2n \), there is a vertical tile in column \( k+1 \).
     - Since there are at most \( n \) vertical tiles, we can now conclude there is a sequence of \( n \) consecutive columns which have no vertical tiles.
     - There are \( 2n \) possible places to place horizontal tiles in these \( n \) columns. If the tiling is maximal, each must be blocked, so there must be a horizontal domino in every row, for \( 2n \) total.
     - Since we also posited the existence of at least one vertical tile in this case, there are at least \( 2n + 1 \) total tiles.

     The same logic shows the same bound if there are at most \( n \) horizontal tiles. The only remaining case is with at least \( n+1 \) tiles of each type, which is \( 2n + 2 \) total tiles and thus too many.

The final answer is \( \boxed{ 2n + 1 } \) for \( n \geq 3 \).