Let $k$ and $n$ be integers such that $k\ge 2$ and $k \le n \le 2k-1$. Place rectangular tiles, each of size $1 \times k$, or $k \times 1$ on a $n \times n$ chessboard so that each tile covers exactly $k$ cells and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.
1. **Case \( n = k \):**
   - When \( n = k \), the \( n \times n \) chessboard is a \( k \times k \) square.
   - We can place \( k \) tiles of size \( 1 \times k \) or \( k \times 1 \) to cover the entire board without any overlap.
   - Therefore, the minimum number of tiles required is \( k \).

2. **Case \( k < n < 2k-1 \):**
   - We need to show that the minimum number of tiles is \( 2(n - k + 1) \).
   - Consider the placement of tiles on the \( n \times n \) board. Each tile covers \( k \) cells.
   - We claim that each row or column can harbor at most one tile since \( n \le 2k - 1 \).
   - Define a row or column as "good" if it contains a tile.
   - **Claim:** If a row or column \( v \) is good, then all rows or columns from the closest side \( S \) to \( v \) are also good.
     - **Proof of Claim:** Suppose \( v \) is a horizontal row and \( S \) is the south side. Consider the set of squares below the tile in \( v \). No vertical tile can touch those squares, else \( S \) would not be the closest side. Thus, if any of those rows was not good, we could place a tile parallel to \( v \), contradicting the assumption that no further tiles can be placed. Hence, all rows from the south side to \( v \) are good. \(\blacksquare\)
   - Let \( a \) be the number of good rows from the north side, \( b \) from the east side, \( c \) from the south side, and \( d \) from the west side.
   - The total number of tiles is \( a + b + c + d \).
   - There is an \((n - a - c) \times (n - b - d)\) rectangle of empty squares. Since it is unfilled, either \( a + c = n \) or \( b + d = n \), or \( n - a - c, n - b - d \le k - 1 \).
   - In the former case, every row (or column) is filled, so there are \( n \) tiles.
   - In the latter case, \( a + b + c + d \ge 2(n - k + 1) \).
   - Therefore, for \( k < n < 2k - 1 \), the minimum number of tiles is \( 2(n - k + 1) \).

3. **Case \( n = 2k - 1 \):**
   - We need to show that the minimum number of tiles is \( 2k - 1 \).
   - Consider the placement of tiles on the \( (2k - 1) \times (2k - 1) \) board.
   - We can place tiles in a staggered manner: left-align a tile in the top row, right-align in the next row, and so on.
   - This configuration ensures that each row and column contains exactly one tile, resulting in \( 2k - 1 \) tiles.

4. **Construction for \( k < n < 2k - 1 \):**
   - Start with an empty \((k - 1) \times (k - 1)\) square.
   - Add four tiles around the perimeter to form a \((k + 1) \times (k + 1)\) square.
   - Place these tiles in the lower-left corner of the \( n \times n \) board.
   - For each of the other \( 2(n - k - 1) \) rows and columns, place a tile in each one.
   - This configuration achieves the value of \( 2(n - k + 1) \).

The final answer is \( \boxed{ k } \) for \( n = k \), \( 2(n - k + 1) \) for \( k < n < 2k - 1 \), and \( 2k - 1 \) for \( n = 2k - 1 \).