When $n = p^{\alpha}, \alpha \ge 2$, arranging them as $p, p^2, \dots, p^{\alpha}$ obviously satisfies the condition.

When $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}, k \ge 3$, first arrange $p_1 p_2 p_3 \cdots p_k p_1$ in order on the circle, then arrange the multiples of $p_j$ between $p_{j-1} p_j$ and $p_j p_{j+1}$. The resulting arrangement satisfies the condition.

Finally, when $n = p^a q^b, a > 1$, first place $pq$ and $p^2 q$ on the circle, put the multiples of $p$ on one arc, and the multiples of $q$ on the other arc.

The remaining case $n = pq$ cannot be achieved.