Find all positive integers $b$ for which there exists a positive integer $a$ with the following properties:

-  $a$ is not a divisor of $b$.

-  $a^a$ is a divisor of $b^b$
To find all positive integers \( b \) for which there exists a positive integer \( a \) with the following properties:
- \( a \) is not a divisor of \( b \).
- \( a^a \) is a divisor of \( b^b \).

We need to analyze the conditions and derive the possible values of \( b \).

1. **Case 1: \( b \) is a prime power**

   Let \( b = p^\alpha \) where \( p \) is a prime and \( \alpha \) is a positive integer. We need to find \( a \) such that:
   - \( a \) is not a divisor of \( b \).
   - \( a^a \) divides \( b^b \).

   If \( a \) is of the form \( p^x \), then:
   - If \( x \ge \alpha \), then \( a^a = (p^x)^a = p^{xa} \) and \( b^b = (p^\alpha)^b = p^{\alpha b} \). Since \( x \ge \alpha \), \( xa \ge \alpha b \), which implies \( a^a > b^b \), a contradiction.
   - If \( x < \alpha \), then \( a \) is a divisor of \( b \), which contradicts the condition that \( a \) is not a divisor of \( b \).

   Therefore, no such \( a \) exists if \( b \) is a prime power.

2. **Case 2: \( b \) has at least two prime divisors**

   Let \( b = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \) where \( k \ge 2 \) and \( p_1 \) is the smallest prime divisor of \( b \).

   - If \( k \ge 3 \) or \( \alpha_2 \ge 2 \) or \( p_2 \ge \frac{(\alpha_1 + 1)p_1}{\alpha_1} \), then \( a = p_1^{\alpha_1 + 1} \) works because \( p_2^{\alpha_2} \cdots p_k^{\alpha_k} \ge \frac{(\alpha_1 + 1)p_1}{\alpha_1} \).

   - The only case left is \( b = p_1^{\alpha_1} p_2 \) where \( \frac{(\alpha_1 + 1)p_1}{\alpha_1} > p_2 \).

     - If \( p_1^{\alpha_1} \ge 2p_2 \), then \( a = p_2^2 \) works.
     - Assume \( p_1^{\alpha_1} < 2p_2 \).

       - If \( \alpha_1 \ge 2 \), then \( 3p_1 \ge \frac{2(\alpha_1 + 1)p_1}{\alpha_1} > 2p_2 > p_1^{\alpha_1} \). Hence, \( 3 > p_1^{\alpha_1 - 1} \). This is true only if \( \alpha_1 = 2 \) and \( p_1 = 2 \). But then \( p_2 < \frac{(\alpha_1 + 1)p_1}{\alpha_1} = 3 \Rightarrow p_2 = 2 \). But \( p_2 > p_1 \), so we get a contradiction. Hence, \( \alpha_1 = 1 \).

       - Then, \( 2p_1 > p_2 \). If there exists such \( a \), then \( p_1^2 \mid a \) or \( p_2^2 \mid a \), but for these values \( a^a \not\mid b^b \). Therefore, for these values of \( b \), such \( a \) does not exist.

Thus, the only values of \( b \) for which such \( a \) does not exist are prime powers and numbers of the form \( pq \) where \( 2p > q > p \) and \( p, q \) are prime numbers.

The final answer is all positive integers except prime powers and numbers in the form \( \boxed{ pq } \) where \( 2p > q > p \) and \( p, q \) are prime numbers.