jmc

Note firstly that $f(i)$ must be an integer, so this means that $i$ must be a perfect square in order for $\sqrt{i}$ to be an integer. Out of the perfect squares, we claim that $i$ must be the square of some prime $p$. For if $\sqrt{i}$ is composite, then it can be written as the product of two integers $a$ and $b$ and we find $f(i)\ge 1+\sqrt{i}+i+a+b>1+\sqrt{i}+i$. Moreover, if $\sqrt{i}$ is prime, then the only factors of $i$ are 1, $\sqrt{i}$, and $i$, so $f(i)=1+\sqrt{i}+i$ as desired. It follows that we only need to calculate the number of primes less than $\sqrt{2010}$. Since $\sqrt{2010}<45$, the desired set of primes is $\{2,3,5,7,11,13,17,19,23,29,31,37,41,43\}$. The set has $\boxed{14}$ elements.