Two positive integers $a$ and $b$ are prime-related if $a = pb$ or $b = pa$ for some prime $p$. Find all positive integers $n$, such that $n$ has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related. 

Note that $1$ and $n$ are included as divisors. 
To solve the problem, we need to find all positive integers \( n \) such that \( n \) has at least three divisors, and all the divisors can be arranged in a circle so that any two adjacent divisors are prime-related. 

1. **Prime Power Case:**
   - If \( n \) is a prime power, say \( n = p^k \) where \( p \) is a prime and \( k \geq 1 \), then the divisors of \( n \) are \( 1, p, p^2, \ldots, p^k \).
   - For \( n \) to have at least three divisors, \( k \geq 2 \).
   - However, in this case, \( 1 \) and \( p \) must be adjacent, and \( p \) and \( p^2 \) must be adjacent. But \( 1 \) and \( p^2 \) are not prime-related, so \( n \) cannot be a prime power.

2. **Perfect Square Case:**
   - If \( n \) is a perfect square, say \( n = m^2 \), then \( n \) has an odd number of divisors.
   - Consider the sum of the exponents of the primes in the prime factorization of a number. Each move (from one divisor to the next) changes the parity of this sum.
   - Starting from \( 1 \) and ending at \( 1 \) after an odd number of moves implies that the parity of the sum of the exponents changes an odd number of times, which is a contradiction.
   - Therefore, \( n \) cannot be a perfect square.

3. **General Case:**
   - We claim that integers \( n \) with at least three prime factors work.
   - Consider \( n = p^a q^b \) where \( p \) and \( q \) are primes and \( a, b \geq 1 \).
   - Assume \( b \) is odd. We can construct a sequence of divisors as follows:
     \[
     1, q, pq, p^2q, \ldots, p^{a-1}q, q^2p^{a-1}, q^2p^{a-2}, \ldots, pq^2, q^2, q^3, q^3p, \ldots, q^b, q^bp, q^bp^2, \ldots, q^bp^{a-1}
     \]
   - This sequence maintains the condition that any two adjacent divisors are prime-related.
   - The remaining divisors are of the form \( p^i \) and \( p^a q^j \) for \( 1 \leq j \leq b \). We attach the following string:
     \[
     p^a q^b, p^a q^{b-1}, \ldots, p^a q, p^a, p^{a-1}, p^{a-2}, \ldots, p
     \]
   - This construction works for \( n = p^a q^b \).

4. **Inductive Step:**
   - Assume that for \( n = \prod_{i=1}^{k-1} p_i^{a_i} \), there is a valid construction.
   - For \( n' = \prod_{i=1}^{k+1} p_i^{a_i} \), consider the sequence:
     \[
     b_1, b_2, \ldots, b_l, p b_l, p, b_{l-1}, \ldots, p b_3, p b_2, p^2 b_2, p^2 b_3, \ldots, p^2 b_l, \ldots
     \]
   - If \( a_{k+1} \) is even, the sequence ends with \( p^{a_{k+1}} b_l \). If \( a_{k+1} \) is odd, it ends with \( p^{a_{k+1}} b_2 \).
   - Add the following string:
     \[
     p^{a_{k+1}} b_1, p^{a_{k+1}-1} b_1, \ldots, p^2 b_1, p b_1
     \]
   - This construction works, and every number which is not a square or a prime power works.

\(\blacksquare\)

The final answer is all positive integers \( \boxed{ n } \) that are not prime powers or perfect squares.