The answer is not. Set the positive real number $a$ and any positive integer $t$. We have $f(a)-f(0)=\sum_{i=0}^{t-1}{f(\frac{(i+1)a}{t})-f(\frac{ia}{t})}>\sum_{i=0}^{t-1}{\sqrt{\frac{a}{t}}=\sqrt{at}}$. Taking large enough $t$ such that $\sqrt{at}>f(a)-f(0)$, we get a contradiction.