Solution:

We say that a positive integer is good if it has the given property. Let $n$ be a good number, and let $d_{1}, d_{2}, \ldots, d_{k}$ be the divisors of $n$ in the circle, in that order. Then for all $1 \leq i \leq k$, $d_{i+1} / d_{i}$ (taking the indices modulo $k$) is equal to either $p_{i}$ or $1 / p_{i}$ for some prime $p_{i}$. In other words, $d_{i+1} / d_{i} = p_{i}^{\epsilon_{i}}$, where $\epsilon_{i} \in \{1, -1\}$. Then
$$p_{1}^{\epsilon_{1}} p_{2}^{\epsilon_{2}} \cdots p_{k}^{\epsilon_{k}} = \frac{d_{2}}{d_{1}} \cdot \frac{d_{3}}{d_{2}} \cdots \frac{d_{1}}{d_{k}} = 1$$
For the product $p_{1}^{\epsilon_{1}} p_{2}^{\epsilon_{2}} \cdots p_{k}^{\epsilon_{k}}$ to equal $1$, any prime factor $p$ must be paired with a factor of $1/p$, and vice versa, so $k$ (the number of divisors of $n$) must be even. Hence, $n$ cannot be a perfect square.

Furthermore, $n$ cannot be the power of a prime (including a prime itself), because $1$ always is a divisor of $n$, and if $n$ is a power of a prime, then the only divisor that can go next to $1$ is the prime itself.

Now, let $n = p^{a} q^{b}$, where $p$ and $q$ are distinct primes, and $a$ is odd. We write the divisors of $n$ in a grid as follows: In the first row, write the numbers $1, q, q^{2}, \ldots, q^{b}$. In the next row, write the numbers $p, p q, p q^{2}, \ldots, p q^{b}$, and so on. The number of rows in the grid, $a+1$, is even. Note that if two squares are adjacent vertically or horizontally, then their corresponding numbers are prime-related. We start with the square with a $1$ in the upper-left corner. We then move right along the first row, move down along the last column, move left along the last row, then zig-zag row by row, passing through every square, until we land on the square with a $p$. The following diagram gives the path for $a=3$ and $b=5$:

Thus, we can write the divisors encountered on this path in a circle, so $n = p^{a} q^{b}$ is good.

Next, assume that $n$ is a good number. Let $d_{1}, d_{2}, \ldots, d_{k}$ be the divisors of $n$ in the circle, in that order. Let $p$ be a prime that does not divide $n$. We claim that $n \cdot p^{e}$ is also a good number. We arrange the divisors of $n \cdot p^{e}$ that are not divisors of $n$ in a grid as follows:
$$\begin{array}{cccc} d_{1} p & d_{1} p^{2} & \ldots & d_{1} p^{e} \\ d_{2} p & d_{2} p^{2} & \ldots & d_{2} p^{e} \\ \vdots & \vdots & \ddots & \vdots \\ d_{k} p & d_{k} p^{2} & \ldots & d_{k} p^{e} \end{array}$$
Note that if two squares are adjacent vertically or horizontally, then their corresponding numbers are prime-related. Also, $k$ (the number of rows) is the number of factors of $n$, which must be even (since $n$ is good). Hence, we can use the same path described above, which starts at $d_{1} p$ and ends at $d_{2} p$. Since $d_{1}$ and $d_{2}$ are adjacent divisors in the circle for $n$, we can insert all the divisors in the grid above between $d_{1}$ and $d_{2}$, to obtain a circle for $n \cdot p^{e}$.

Finally, let $n$ be a positive integer that is neither a perfect square nor a power of a prime. Let the prime factorization of $n$ be
$$n = p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{t}^{e_{t}}$$
Since $n$ is not the power of a prime, $t \geq 2$. Also, since $n$ is not a perfect square, at least one exponent $e_{i}$ is odd. Without loss of generality, assume that $e_{1}$ is odd. Then from our work above, $p_{1}^{e_{1}} p_{2}^{e_{2}}$ is good, so $p_{1}^{e_{1}} p_{2}^{e_{2}} p_{3}^{e_{3}}$ is good, and so on, until $n = p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{t}^{e_{t}}$ is good.

Therefore, a positive integer $n$ has the given property if and only if it is neither a perfect square nor a power of a prime.