Determine all positive integers $n$ such that all positive integers less than $n$ and coprime to $n$ are powers of primes.
1. **Prime Factorization and Setup:**
   Let \( n = \prod_{i=1}^{\infty} p_i^{a_i} \), where \( p_i \) is the \( i \)-th prime and \( a_i \) are nonnegative integers. We need to find all positive integers \( n \) such that all positive integers less than \( n \) and coprime to \( n \) are powers of primes.

2. **Identifying \( r \) and \( s \):**
   Let \( r \) be the smallest positive integer for which \( a_r = 0 \) and let \( s \) be the second smallest positive integer for which \( a_s = 0 \). This means \( p_r \) and \( p_s \) are the smallest primes not dividing \( n \).

3. **Fundamental Inequality:**
   Since \( \gcd(p_r p_s, n) = 1 \), we must have \( p_r p_s > n \). Additionally, we have:
   \[
   n \ge \frac{1}{p_r} \prod_{i=1}^{s-1} p_i
   \]
   Combining these, we get:
   \[
   p_r p_s > \frac{1}{p_r} \prod_{i=1}^{s-1} p_i
   \]
   Simplifying, we obtain:
   \[
   p_r^2 p_s > \prod_{i=1}^{s-1} p_i
   \]

4. **Using Bertrand's Postulate:**
   Bertrand's postulate states that for any integer \( k \ge 2 \), there is always at least one prime \( p \) such that \( k < p < 2k \). Using this, we can show that the inequality \( p_r^2 p_s > \prod_{i=1}^{s-1} p_i \) holds only for small values of \( s \).

5. **Bounding \( s \):**
   By checking the inequality for small values of \( s \), we find that it holds only when \( s \le 5 \). This means \( s \) can be at most 5.

6. **Checking for Larger \( t \):**
   Assume there exists an integer \( t > s \) such that \( a_t \ne 0 \). Then we would need:
   \[
   p_r p_s > p_{s+1} \prod_{i=1}^{s-1} p_i
   \]
   Using Bertrand's postulate and some quick multiplication, we find that this can only occur if \( n = 5 \) or \( n = 14 \).

7. **Checking All Cases:**
   Since \( n < p_r p_s \le p_4 p_5 = 77 \), we need to check all integers \( n \) from 1 to 76. After checking, we find that the solutions are:
   \[
   n \in \{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42\}
   \]

The final answer is \( \boxed{n \in \{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42\}} \)