Saudi Arabian Mathematical Competitions, 2012

The inequality must hold for $x = n$. Thus
$$1 + \sqrt{2} + \dots + \sqrt{n-1} < n.$$
But $1 + \sqrt{2} + \sqrt{3} > 4$, and by induction
$$1 + \sqrt{2} + \dots + \sqrt{n-1} > n \quad \text{for all } n \ge 4.$$
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 $n = 2$.