62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Suppose that $x_i=x_{i+1}$ for some $i$ (from now on we denote $x_{n+k}=x_k$.) Then from the equality $x_i-\frac{1}{x_{i+1}}=x_{i+1}-\frac{1}{x_{i+2}}$ we get $x_{i+1}=x_{i+2}$, then $x_{i+2}=x_{i+3}$ and so on. So we will get that all numbers are equal.

Suppose now that we don't have equal adjacent numbers. From the statement, we get $$x_i-x_{i+1}=\frac{1}{x_{i+1}}-\frac{1}{x_{i+2}}=-\frac{x_{i+1}-x_{i+2}}{x_{i+1}{x}_{i+2}},i=\underline{1,n}.$$ If we multiply all such equalities, the product of the differences will cancel out, and we will get $\frac{1}{(x_1{x}_2\dots x_n{)}^2}=(-1{)}^n$. For an odd $n$ it's impossible, and for even $n$ one possible array is $(2,\frac{1}{2},2,\frac{1}{2},\dots ,2,\frac{1}{2})$.