Saudi Arabia Mathematical Competitions 2012

The inequality must hold for $x=n$. Thus $$1+\sqrt{2}+\cdots +\sqrt{n-1}<n.$$ But $1+\sqrt{2}+\sqrt{3}>4$, and by induction $$1+\sqrt{2}+\cdots +\sqrt{n-1}>n\text{for all }n\ge 4.$$ Therefore, it is necessary that $n\le 3$.

When $n=3$, taking $x=4$ also gives $\sqrt{3}+\sqrt{2}+1>4$, not possible.

When $n=2$, we have $$\sqrt{x-1}+\sqrt{x-2}<2\sqrt{x-1}\le x,$$ $$\text{since we have }{x}^2-4x+4=(x-2{)}^2\ge 0.$$ When $n=1$, we have $\sqrt{x-1}<x$, since it is equivalent to $$x^2-x+1>0.$$ The positive integers $n$ satisfying the property are $n=1$ and $n=2$.