Determine all positive integers $n$ such that the number $n(n+2)(n+4)$ has at most $15$ positive divisors.
1. **Understanding the problem:**
   We need to find all positive integers \( n \) such that the number \( n(n+2)(n+4) \) has at most 15 positive divisors. 

2. **Using the divisor function \(\tau\):**
   The number of divisors of a number \( n \) is denoted by \(\tau(n)\). For any two coprime numbers \( m \) and \( n \), \(\tau(mn) = \tau(m) \tau(n)\).

3. **Case 1: \( n \) is odd:**
   - If \( n \) is odd, then \( n, n+2, n+4 \) are pairwise coprime.
   - Therefore, \(\tau(n(n+2)(n+4)) = \tau(n) \tau(n+2) \tau(n+4)\).
   - Given that \(\tau(n(n+2)(n+4)) \leq 15\), we have \(\tau(n) \tau(n+2) \tau(n+4) \leq 15\).

4. **Analyzing \(\tau(n)\), \(\tau(n+2)\), and \(\tau(n+4)\):**
   - Since \(\tau(n) \geq 2\) for any positive integer \( n \), the possible values for \(\tau(n)\), \(\tau(n+2)\), and \(\tau(n+4)\) are from the set \(\{2, 3\}\).
   - There can be at most one 3 among \(\tau(n)\), \(\tau(n+2)\), and \(\tau(n+4)\).

5. **Considering \(\tau(n) = 3\):**
   - If \(\tau(n) = 3\), then \( n \) must be a square of a prime, say \( n = p^2 \).
   - We need \( n+2 = p^2 + 2 \) to be prime. Checking modulo 3, we find that \( n = 9 \) is the only such number.
   - Similarly, if one of \( n, n+2, n+4 \) is a square of a prime and the other two are primes, then \( n = 9, n = 7, n = 5 \) are the only solutions.

6. **Considering all three numbers are primes:**
   - If all three numbers are primes, then \( n = 3 \) is the only solution, taking modulo 3 again.

7. **Conclusion for odd \( n \):**
   - The only odd \( n \) satisfying the conditions are \( \boxed{1, 3, 5, 7, 9} \).

8. **Case 2: \( n \) is even:**
   - Let \( n = 2k \) for some positive integer \( k \).
   - Then \( n(n+2)(n+4) = 8k(k+1)(k+2) \).
   - Since \( k, k+1, k+2 \) are pairwise coprime, we proceed similarly to the previous case.
   - We must have \( k = 1, 2 \), which gives the only even solutions of \( n \) to be \( \boxed{2, 4} \).

9. **Combining both cases:**
   - The complete set of solutions is \( \boxed{1, 2, 3, 4, 5, 7, 9} \).