Let's prove this statement by induction on the number of distinct prime divisors of a given number. In this case, this number will be denoted by $n$, the number of its divisors — by $d(n)$ and the divisors written in counterclockwise order — by $a_1, a_2, \dots, a_{d(n)}$, where $a_1 = 1$.

Base: let $n = p^k$, where $p$ is a prime number. Let's arrange all divisors of the number $n$ in ascending order: $a_1 = 1, a_2 = p, \dots, a_{k+1} = p^k$. It is obvious that such arrangement satisfies the condition.

Step: suppose the assertion is proved for all numbers having at most $s$ different prime divisors, let us prove it for an arbitrary number $n$ having $s + 1$ prime divisors. We write $n$ as $n = p^k m$, where $p$ is a prime number that does not divide $m$.

If $\ell$ is a sequence of numbers written in a clockwise circle then by $\ell^*$ we will denote the same sequence but written in reverse order, i. e., if $\ell$ is $a_1, a_2, \dots, a_k$, then $\ell^*$ is $a_k, \dots, a_2, a_1$. Moreover by $q\ell$ we denote the sequence obtained from $\ell$ by multiplying all its terms by $q$, i. e. $q\ell$ is $qa_1, qa_2, \dots, qa_k$.

By the induction hypothesis all divisors of the number $m$ can be arranged in a circle according to the condition: $a_1, a_2, \dots, a_{d(m)}$. Denote this sequence $\ell$. Then we arrange the divisors of the number $n$ around the circle in the following order:

$\ell, (p\ell)^*, p^2\ell, \dots, (p^{k-1}\ell)^*, p^k\ell$, if $k$ is odd,
and
$\ell, (p\ell)^*, p^2\ell, \dots, p^{k-1}\ell, (p^k\ell)^*$, if $k$ is even.

It is easy to see that these sequences satisfy the condition of the problem.