Find all positive integers $n=p_1p_2 \cdots p_k$ which divide $(p_1+1)(p_2+1)\cdots (p_k+1)$ where $p_1 p_2 \cdots p_k$ is the factorization of $n$ into prime factors (not necessarily all distinct).
1. **Assume \( p_k \) is the largest prime factor in the set \(\{p_1, p_2, \ldots, p_k\}\).**
   - Since \( p_k \) is the largest prime factor, we have \( p_k \mid (p_1+1)(p_2+1)\cdots(p_k+1) \).
   - This implies \( p_k \mid p_j+1 \) for some \( j \).

2. **Consider the case when \( p_k > 3 \):**
   - If \( p_k > 3 \), then \( p_j > 2 \) because \( p_k \) is the largest prime factor.
   - Since \( p_j \) is a prime number greater than 2, \( p_j+1 \) must be composite.
   - Therefore, \( p_k < p_j+1 \), which implies \( p_k \leq p_j \).
   - This leads to \( p_k = p_j \), which contradicts the fact that \( p_k \mid p_j+1 \) because a prime number cannot divide itself plus one.

3. **Conclude that \( p_k = 3 \):**
   - Since \( p_k \) cannot be greater than 3, the only possible value for \( p_k \) is 3.

4. **Assume there are \( r \geq 0 \) numbers \( p_j \) equal to 2 and \( s = k - r \geq 0 \) numbers equal to 3:**
   - We need to check if \( n = 2^r \cdot 3^s \) divides \( (2+1)^r \cdot (3+1)^s = 3^r \cdot 4^s \).

5. **Check the divisibility condition \( 2^r \cdot 3^s \mid 3^r \cdot 4^s \):**
   - Rewrite \( 4^s \) as \( (2^2)^s = 2^{2s} \).
   - Therefore, the condition becomes \( 2^r \cdot 3^s \mid 3^r \cdot 2^{2s} \).

6. **Analyze the powers of 2 and 3:**
   - For the powers of 2: \( 2^r \mid 2^{2s} \) implies \( r \leq 2s \).
   - For the powers of 3: \( 3^s \mid 3^r \) implies \( s \leq r \).

7. **Combine the inequalities:**
   - We have \( s \leq r \leq 2s \).

8. **Conclusion:**
   - The positive integers \( n \) that satisfy the given condition are of the form \( n = 2^r \cdot 3^s \) where \( 0 \leq s \leq r \leq 2s \).

The final answer is \( \boxed{ n = 2^r \cdot 3^s } \) where \( 0 \leq s \leq r \leq 2s \).