Find all positive integers $n$, for which there exists a positive integer $k$ such that for every positive divisor $d$ of $n$, the number $d - k$ is also a (not necessarily positive) divisor of $n$.
To find all positive integers \( n \) for which there exists a positive integer \( k \) such that for every positive divisor \( d \) of \( n \), the number \( d - k \) is also a (not necessarily positive) divisor of \( n \), we proceed as follows:

1. **Case \( n = 3 \):**
   - Let \( n = 3 \) and \( k = 2 \).
   - The divisors of \( n = 3 \) are \( 1 \) and \( 3 \).
   - For \( d = 1 \), \( d - k = 1 - 2 = -1 \), which is not a divisor of \( 3 \).
   - For \( d = 3 \), \( d - k = 3 - 2 = 1 \), which is a divisor of \( 3 \).
   - Therefore, \( n = 3 \) does not satisfy the condition for \( k = 2 \).

2. **General Case \( n > 3 \):**
   - Suppose \( n > 3 \). We need \( k - 1 \mid n \) and \( n - k \mid n \).
   - Let \( d \) be a divisor of \( n \). Then \( d - k \) must also be a divisor of \( n \).
   - Consider the maximum of \( k - 1 \) and \( n - k \). We have:
     \[
     \max(k - 1, n - k) \geq \frac{n - 1}{2} > \frac{n}{3}
     \]
   - This implies \( n = 2 \max(k - 1, n - k) \) or \( n = k - 1 \).

3. **Subcase \( n = 2k - 2 \):**
   - If \( n = 2k - 2 \), then \( k - 2 \mid 2k - 2 \).
   - This simplifies to \( k - 2 \mid 2 \), so \( k - 2 \) can be \( 1 \) or \( 2 \).
   - If \( k - 2 = 1 \), then \( k = 3 \) and \( n = 4 \).
   - If \( k - 2 = 2 \), then \( k = 4 \) and \( n = 6 \).

4. **Subcase \( n = 2k \):**
   - If \( n = 2k \), then \( k = n + 1 \), which is a contradiction because \( 0 \) should be a divisor.

5. **Prime \( n \):**
   - Let \( d > 1 \) and \( d \leq \frac{n}{2} \) be a divisor of \( n \).
   - Then \( n + 1 - d \mid n \) implies \( n + 1 - d \leq \frac{n}{2} \).
   - This leads to \( n + 2 \leq 2d \), which is a contradiction.
   - Therefore, \( n \) must be prime.

The final answer is \( \boxed{ n = 3, 4, 6 } \) or prime.