We will denote the floor of $x$ as $[x]$. Note that each number $d$ in $1, \ldots, m$ appears $\left [ \frac{m}{d} \right ]$ times in $\sum_{i=1}^m f(i)$. That means we have \begin{align*} \sum_{i=1}^m f(i) &= \sum_{d=1}^m \frac{1}{1+d} \left [ \frac{m}{d} \right ] \\ &\leq \sum_{d=1}^m \frac{1}{1+d} \frac{m}{d}\\ &=\sum_{d=1}^m \frac{m}{d} - \frac{m}{1+d} \\ &=m-\frac{m}{m+1} < m \end{align*}