Solve $\begin{cases} \left|x_1-x_2\right|=\left|x_2-x_3\right|=...=\left|x_{2018}-x_1\right|,\\ x_1+x_2+...+x_{2018}=2018. \end{cases}$
I think there must be such a way to solve systems of equation with the form of rotation and absolute value like this.
I have difficulty in solving the first equation. Since to me, there are quite a lot of cases to consider, for example, $\left|x_1-x_2\right|=\left|x_2-x_3\right|$ leads to $x_1-x_2=x_2-x_3$ and $x_1-x_2=x_3-x_2$ and so on.
Here is the answer to verify my comment.
Let $|x_{1} - x_{2}| = C \geq 0$ and note that \begin{align} |x_{1} - x_{2}| = C \implies x_{1} = \pm C + x_{2} \end{align} Similarly, we have \begin{align} x_{1} &= \pm C + x_{2} \\ x_{2} &= \pm C + x_{3} \\ & \; \; \vdots \\ x_{2017} &= \pm C + x_{2018} \\ x_{2018} &= \pm C + x_{1} \\ \end{align} Keep substituting, we obtain \begin{align} x_{1} &= \pm C + x_{2} \\ &= \pm C \pm C + x_{3} \\ & \; \; \vdots \\ &= \pm C \pm C \pm \cdots \pm C + x_{1} \end{align} Thus, we have \begin{align} \pm 2018C = 0 \implies C = 0 \end{align} It means that $x_{1} = x_{2} \cdots = x_{2018}$ and let them be $A$.
So now, we use the second equation \begin{align} x_{1} + x_{2} + \cdots + x_{2018} &= 2018 \\ 2018A &= 2018 \\ A &= 1 \end{align}
Conclusion: $x_{1} = x_{2} \cdots = x_{2018} = 1$