We always take indices modulo $n$; this also means that if $i=0$ and we refer to $\dfrac{x_i}{i} = \dfrac{x_0}{0}$, then we actually refer to $\dfrac{x_n}{n}$, and so on. Note that $(x_1, x_2, \ldots, x_n) = (1,1,\ldots,1)$ is a solution. Also note that $x_i \ge 1$ for all $i$ (from $\max\{i-1, x_{i-1}\} = (i-1)x_i$, we have $(i-1)x_i \ge i-1$ and so $x_i \ge 1$), so any other solution has $x_a > 1$ for some $a$. Now see how this propagates through the other variables. $x_a > 1$. If $x_a \le a$, then $\max{a, x_a} = a$, so $x_{a+1} = x_{a+2} = \ldots = x_n = x_1 = \ldots = x_{a-1} = 1$, so $\max\{a-1, x_{a-1}\} = a-1$ and so $x_a = 1$, contradiction. So $x_a > a$ and so $x_{a+1} = \dfrac{x_a}{a} > 1$. We can induct to prove that $x_i > i$ for all $i$, so $x_i = \dfrac{x_{i-1}}{i-1}$ for all $i$. So $x_a = \dfrac{x_a}{(a-1)(a-2)(a-3)\ldots(1)(n)(n-1)\ldots(a)}$, so $x_a = x_a \cdot n!$. Because $n \ge 2$, we have $x_a = 0$, contradiction. So the only solution is $(x_1, x_2, \ldots, x_n) = (1,1,\ldots,1)$.