National Math Olympiad - Final Round, 2013

Let $y$ be such a number. Then $\sqrt{x} \ge |\sqrt{x}| = y$. Since $\sqrt{x} \ge 0$ we have $y = [\sqrt{x}] \ge 0$, so we may square the inequality to get $x \ge y^2$. Also, $\frac{x+23}{8} < [\frac{x+23}{8}]+1 = y+1$, or $x < 8y-15$. This implies $y^2 < 8y-15$, or $(y-3)(y-5) < 0$, which means that $3 < y < 5$. But $y$ is an integer, so $y=4$. In this case there indeed exists a real number $x$ which satisfies the condition, for example $x=16$.