Using QM-Am we can write: $\sqrt{\frac{{{\sum{\left| {{a}_{i}}-{{a}_{i+1}} \right|}}^{2}}}{n}}\ge \frac{\sum{\left| {{a}_{i}}-{{a}_{i+1}} \right|}}{n}\Rightarrow \sum{\left| {{a}_{i}}-{{a}_{i+1}} \right|}\le \sqrt{2n}\sqrt{1-\sum{{{a}_{i}}{{a}_{i+1}}}}\le \sqrt{2n}$ but I am not sure if we can have or not ${{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+...+{{a}_{n-1}}{{a}_{n}}+{{a}_{n}}{{a}_{1}}=0$

Case 1: $2\mid n$ $\begin{aligned} 0\le \sum_{i=1}^{n}(a_i+a_{i+1})^2=2+2\sum_{i=1}^{n}a_ia_{i+1} \Rightarrow \sum_{i=1}^{n}a_ia_{i+1}\ge -1 \end{aligned}$ $\begin{aligned} (|a_1-a_2|+|a_2-a_3| +\dots +|a_n-a_1|)^2 & \le n(\sum_{i=1}^{n}|a_i-a_{i+1}|^2)\\ & =n(2-2\sum_{i=1}^{n}a_ia_{i+1})\\ & \le n(2+2)\\ & =4n \end{aligned}$ $\begin{aligned} \Rightarrow |a_1-a_2|+|a_2-a_3| +\dots +|a_n-a_1|\le 2\sqrt{n} \end{aligned}$ When $a_1=a_3=\cdots =a_{n-1}=\frac{1}{\sqrt{n}}, a_2=a_4=\cdots =a_{n}=-\frac{1}{\sqrt{n}}$, the equality can occur. So the maximum value of this case is $2\sqrt{n}$ Case 2: $2\nmid n$ Then we know that there is a subscript $i\in \{1,2,\cdots ,n \}$ such that $(a_{i-1}-a_i)(a_i-a_{i+1})\ge 0$ (Otherwise, we have $(a_{i-1}-a_i)(a_i-a_{i+1})< 0,i=1,2,\cdots, n$ $\Rightarrow 0> \prod_{i=1}^{n}((a_{i-1}-a_i)(a_i-a_{i+1}))=\prod_{i=1}^{n}(a_{i-1}-a_i)^2$, contradiction) $WLOG$ $i=n$, then $(a_{n-1}-a_n)(a_n-a_1)\ge 0$ $\Rightarrow |a_{n-1}-a_n|+|a_n-a_1|=|a_{n-1}-a_1|$ $\begin{aligned} |a_1-a_2|+|a_2-a_3| +\dots +|a_n-a_1| & =|a_1-a_2|+|a_2-a_3| +\dots +|a_{n-1}-a_1| \end{aligned}$ $0\le (a_1+a_2)^2+(a_2+a_3)^2+\cdots +(a_{n-1}+a_1)^2=2\sum_{i=1}^{n-1}a_i^2+2(a_1a_2+a_2a_3+\cdots +a_{n-1}a_1)$ $\Rightarrow a_1a_2+a_2a_3+\cdots +a_{n-1}a_1\ge -\sum_{i=1}^{n-1}a_i^2\ge -1$ $\begin{aligned} (|a_1-a_2|+|a_2-a_3| +\dots +|a_n-a_1|)^2 & =(|a_1-a_2|+|a_2-a_3| +\dots +|a_{n-1}-a_1|)^2\\ & \le (n-1)((a_1-a_2)^2+(a_2-a_3)^2 +\dots +(a_{n-1}-a_1)^2)\\ & =(n-1)(2\sum_{i=1}^{n-1}a_i^2-2(a_1a_2+a_2a_3+\cdots +a_{n-1}a_1))\\ & \le (n-1)(2+2)\\ & =4(n-1) \end{aligned}$ $\begin{aligned} \Rightarrow |a_1-a_2|+|a_2-a_3| +\dots +|a_n-a_1|\le 2\sqrt{n-1} \end{aligned}$ When $a_1=a_3=\cdots =a_{n-2}=\frac{1}{\sqrt{n-1}}, a_2=a_4=\cdots =a_{n-1}=-\frac{1}{\sqrt{n-1}},a_n=0$, the equality can occur. So the maximum value of this case is $2\sqrt{n-1}$