THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

It is known that $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{(n-1)n}+\frac{1}{n}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots +\left(\frac{1}{n-1}-\frac{1}{n}\right)+\frac{1}{n}=1.$$ If $n$ is not of the form $k(k+1)$, $k\in \mathbb{N}$, then the numbers $x_1=2$, $x_2=2\cdot 3,\dots$, $x_{n-1}=(n-1)n$, $x_n=n$ are pairwise distinct, therefore they fulfill the conditions.

If $n=k(k+1)$, $k\in \mathbb{N}$ (the smallest $n\ge 3$ of this form is $n=6$), then the equality $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{(n-2)(n-1)}+\frac{1}{n-1}=1$$ can be written $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{(n-3)(n-2)}+\frac{1}{(n-2)(n-1)+1}+\frac{1}{(n-2)(n-1)[(n-2)(n-1)+1]}+\frac{1}{n-1}=1.$$ The numbers $x_1=2$, $x_2=2\cdot 3,\dots$, $x_{n-3}=(n-3)(n-2)$, $x_{n-2}=(n-2)(n-1)+1$, $x_{n-1}=(n-2)(n-1)[(n-2)(n-1)+1]$, $x_n=n-1$ are pairwise different. Moreover, $n\le (n-3)(n-2)$, $\forall n\ge 6$, hence $n\in \{x_2,x_3,\dots ,x_{n-3}\}$, therefore the numbers $x_1,x_2,\dots ,x_n$ fulfill all the conditions.