Junior Balkan Mathematical Olympiad

a) If $x_i=x_{i+1}$ for some $i$ (assuming $x_{n+1}=x_1$), then by the given identity all $x_i$ will be equal, a contradiction. Thus $x_1\ne x_2$ and $$x_1-x_2=k\frac{x_2-x_3}{x_2{x}_3}.$$ Analogously $$x_1-x_2=k\frac{x_2-x_3}{x_2{x}_3}=k^2\frac{x_3-x_4}{(x_2{x}_3)(x_3{x}_4)}=\cdots =k^n\frac{x_1-x_2}{(x_2{x}_3)(x_3{x}_4)\dots (x_1{x}_2)}.$$ Since $x_1\ne x_2$ we get $$x_1{x}_2\dots x_n=\pm \sqrt{k^n}=\pm k^{\frac{n-1}{2}}\sqrt{k}.$$ If among these two values, positive or negative, is obtained, then the other one will be also obtained by changing the sign of all $x_i$ since $n$ is odd.

b) From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k\ge 4$. For $k=4$ let $n=2019$ and $x_{3j}=4$, $x_{3j-1}=1$, $x_{3j-2}=-2$ for $j=1,2,\dots ,673$. So, the required least value is $k=4$.