Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
1. **Expression Analysis**:
   Given \( P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9) \), we need to find the largest integer that is a divisor of \( P(n) \) for all positive even integers \( n \).

2. **Oddness of \( P(n) \)**:
   Since \( n \) is even, each term \( n + k \) (where \( k \) is odd) will be odd. Therefore, \( P(n) \) is the product of five odd numbers, making \( P(n) \) itself odd.

3. **Prime Factors**:
   To find the largest integer that divides \( P(n) \) for all even \( n \), we need to consider the prime factors that will always appear in the product \( P(n) \).

4. **Consecutive Terms**:
   Consider the expressions for \( P(n) \) and \( P(n+2) \):
   \[
   P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)
   \]
   \[
   P(n + 2) = (n + 3)(n + 5)(n + 7)(n + 9)(n + 11)
   \]
   Notice that each term within the brackets will appear in five consecutive even \( n \).

5. **Prime Divisors**:
   We need to find the primes that will always appear in the product of any five consecutive odd numbers. The primes less than or equal to 5 are 2, 3, and 5. However, since \( P(n) \) is always odd, 2 is not a factor. We need to check for 3 and 5.

6. **Divisibility by 3**:
   Among any five consecutive odd numbers, at least one of them is divisible by 3. This is because every third odd number is divisible by 3.

7. **Divisibility by 5**:
   Similarly, among any five consecutive odd numbers, at least one of them is divisible by 5. This is because every fifth odd number is divisible by 5.

8. **Combining Factors**:
   Since 3 and 5 are always factors of \( P(n) \), the product \( 3 \times 5 = 15 \) will always divide \( P(n) \).

Conclusion:
The largest integer that is a divisor of \( P(n) \) for all positive even integers \( n \) is \( \boxed{15} \).