Problem 867

4. Let $a_{18}=a_{1}$. Denote
$$S=\left(a_{1}-a_{2}\right)\left(a_{2}-a_{3}\right) \cdots\left(a_{16}-a_{17}\right)\left(a_{17}-a_{1}\right) \text {. }$$

Proceed in the following four steps:
(1) $n$ is even.

Since $a_{1}, a_{2}, \cdots, a_{17}$ contain nine odd numbers and eight even numbers, there must exist $i \in\{1,2, \cdots, 17\}$ such that both $a_{i}$ and $a_{i+1}$ are odd. Hence, $a_{i}-a_{i+1}$ is even. Therefore, $n$ is even.
\begin{array}{l} \text { (2) } na_{i+1}\right\}, \\ J=\left\{j \mid j \in\{1,2, \cdots, 17\}, a_{j}0$. Thus, $\varphi(x)$ is monotonically decreasing on the interval $(0,8.5)$ and monotonically increasing on the interval $(8.5,17)$. Therefore, when $p \in\{1,2, \cdots, 16\}$,
p^{p}(17-p)^{17-p}=\mathrm{e}^{\varphi(p)}
has a minimum value of $8^{8} \times 9^{9}$. Hence, $S \leqslant \frac{72^{17}}{8^{8} \times 9^{9}}=8^{9} \times 9^{8}<10^{17}$. Therefore, $n<10$. (3) $n \neq 8$. Let $t \in\{1,2,3,4\}$. Clearly, the remainders of $a_{1}, a_{2}, \cdots, a_{17}$ modulo $2^{t}$ cover 0,1, $\cdots, 2^{t}-1$. For $i=1,2, \cdots, 17$, when $a_{i}$ and $a_{i+1}$ have different remainders modulo $2^{t}$ (there are at least $2^{t}$ such $i$), $2^{t} \times\left(a_{i}-a_{i+1}\right)$. Thus, there are at most $17-2^{t}$ values of $i \in\{1,2, \cdots, 17\}$ such that $2^{t}$ divides $\left(a_{i}-a_{i+1}\right)$. In particular, among $a_{i}-a_{i+1} (i=1,2, \cdots, 17)$, there are at most 15 even numbers, of which at most 13 are multiples of $2^{2}$, at most 9 are multiples of $2^{3}$, and at most 1 is a multiple of $2^{4}$. Thus, the number of times the prime factor 2 appears in $S=\prod_{i=1}^{17}\left(a_{i}-a_{i+1}\right)$ is at most $15+13+9+1=38$. If $n=8$, then $S=8^{17}=2^{51}$, which is a contradiction. Therefore, $n \neq 8$. (4) Construct a permutation $a_{1}, a_{2}, \cdots, a_{17}$ of $1,2, \cdots, 17$ such that $S=6^{17}$. Let $a_{1}, a_{2}, \cdots, a_{17}$ be $1,9,17,8,7,16$, $14,5,13,4,6,15,12,3,11,2,10$. Then,
\begin{aligned}
S= & (-8)(-8) \times 9 \times 1 \times(-9) \times 2 \times \\
& 9 \times(-8) \times 9 \times(-2)(-9) \times 3 \times \\
& 9 \times(-8) \times 9 \times(-8) \times 9 \\
= & (-1)^{8} 2^{3+3+1+3+1+3+3} \times \\
& 3^{2+2+2+2+2+1+2+2+2} \\
= & 6^{17} .
\end{aligned}
$$$$

Combining (1) to (4), we conclude that the maximum value of $n$ is 6.