Let $n$ and $k$ be positive integers such that $\frac{1}{2}n < k \leq \frac{2}{3}n$. Our goal is to find the least number $m$ for which it is possible to place $m$ pawns on an $n \times n$ chessboard such that no column or row contains a block of $k$ adjacent unoccupied squares.

### Analysis:
1. Chessboard Structure:
The chessboard is $n \times n$, meaning it contains $n^2$ squares.

2. Occupied Squares and Condition:
The condition states that in any row or column, there should not be $k$ consecutive unoccupied squares. Hence, each row or column must be interrupted by pawns frequently to avoid these consecutive spaces.

### Strategy to Satisfy the Conditions:
The strategy is to divide each row and each column evenly with pawns such that no $k$ adjacent unoccupied squares appear.

### Steps to Place Pawns:
1. Rows Consideration:
- For each row of length $n$, identify the longest possible block of consecutive squares we can have without placing a pawn, which is $< k$.

2. Calculation for Rows:
- If there are $n$ squares in a row, we need to ensure that every possible group of $k$ or more squares has at least one pawn.
- If $d = n - k$, the placement of pawns should break the row such that each break occurs before a block gets as long as $k$. This requires a pawn after every $k - 1$ squares.
- Number of necessary divisions (pawns) in a row = $\left\lceil \frac{n}{k} \right\rceil$.
- Maximum possible value of these divisions due to edge adjustments will lead to $2(n-k)$, accounting for rows and the similar applies to columns.

3. Effective Placement:
- Optimize so that if each row is broken into $\left\lceil \frac{n}{k} \right\rceil$ blocks, ensuring all adjustments for bordering overlaps lead to the conclusion of needing $4(n-k)$ pawns in total, considering both rows and columns.

Therefore, the least number $m$ of pawns needed to ensure the chessboard satisfies the conditions is:

$$\boxed{4(n-k)}$$