smc

For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square. Since $\sqrt{x}$ can't be negative, $120-\sqrt{x}\le 120$. The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$, so there are $11$ values for $120-\sqrt{x}$. Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$, the number of values of $x$ is $\boxed{11\text{.}}$.