Of course $n \in \mathbb{Z_+}$ .

$n_{\text{max}}=6$ try to prove $n＜10$ and get a constraction for $n=8$ such as $v_2\left(\prod_{i=1}^17{(a_i-a_{i+1}})\right)≤38$

It is not difficult to show that $n$ is even. We define $\left| a_i-a_{i+1} \right|=x_i-y_i$, where $x_i=\max\left\{ a_i, a_{i+1} \right\}$ and $y_i=\min\left\{ a_i, a_{i+1} \right\}$. It is clear that each $a_k$ can appear at most twice among the numbers $x_1$, $x_2$, ... , $x_{17}$, and it can also appear at most twice among the numbers $y_1$, $y_2$, ... , $y_{17}$, then $$\sum x_i-\sum y_i\leq \left [ 2(17+16+\cdots +10)+9 \right ]-\left [ 2(1+2+\cdots +8)+9 \right ]=144.$$For inequality of means we have that $$n\leq \left| n \right|=\sqrt[17]{\prod_{i=1}^{17}\left| a_i-a_{i+1} \right|}\leq \frac{\sum x_i-\sum y_i}{17}\leq \frac{144}{17}.$$Therefore, $n\leq 8$. If $n=8$, then $$(a_1-a_2)(a_2-a_3)\cdots (a_{17}-a_1)=8^{17}.$$All $17$ factors cannot be multiples of $8$, otherwise all $17$ $a_1$, $a_2$, ... , $a_{17}$ numbers would have the same parity and that is false. Therefore, at least one of the $17$ factors is a multiple of $16$. This is only possible if the numbers in that factor are $1$ and $17$. Of the remaining $16$ factors, only $9$ can be multiples of $8$: $$(17, 9), (16, 8), (15, 7), (14, 6), (13, 5), (12, 4), (11, 3), (10, 2) \hspace{0.1cm}\text{and}\hspace{0.1cm} (9, 1)$$Therefore, $$51=3\cdot 17=v_2\left ( \prod_{i=1}^{17}(a_i-a_{i+1}) \right )\leq 4+9\cdot 3+7\cdot 2=45,$$which is false. This means that $n\leq 6$. The example is for $n=6$ is: $$(17-8)(8-16)(16-7)(7-15)(15-6)(6-14)(14-5)(5-13)(13-4)(4-2)(2-11)(11-10)(10-1)(1-3)(3-12)(12-9)(9-17)=6^{17}.$$