, 2021

We prove by induction. Let $a = \max S$ and use the inductive hypothesis to find an ordering $b_1, b_2, \dots, b_{n-1}$ of $S \setminus \{a\}$ so that $b_i$ does not divide $b_{i-1} \pm b_{i+1}$ for all $2 \le i \le n-1$. Observe that
$$|b_j - b_{j+1}| < a \neq b_j + b_{j+1} < 2a,$$
so that $a$ divides neither $b_j - b_{j+1}$ nor $b_j + b_{j+1}$. Thus if $b_1, b_2, \dots, b_j, a, b_{j+1}, \dots, b_{n-1}$ does not satisfy the desired property, then either $b_j \mid a + \epsilon b_{j-1}$ or $b_{j+1} \mid a + \epsilon b_{j+2}$ for some $\epsilon \in \{1, -1\}$. Since there are $n$ positions to insert $a$ and only $n-1$ of $b_j$, at least one $b_j$ “violates the condition twice as the divider”, meaning $b_j \mid a + \epsilon_1 b_{j-1}$ and $b_j \mid a + \epsilon_2 b_{j+1}$ for some $\epsilon_1, \epsilon_2 \in \{1, -1\}$. Thus, $\epsilon_2 b_{j+1} \equiv -a \equiv \epsilon_1 b_{j-1} \pmod{b_j}$, contradicting the induction hypothesis.