Solution:
The two equations are equivalent to $2017\lfloor x\rfloor=2017$ and $\{2017 x\}=\frac{1}{2017}$, respectively. The first equation reduces to $\lfloor x\rfloor=1$, so we must have $x=1+r$ for some real $r$ satisfying $0 \leq r<1$. From the second equation, we deduce that $\{2017 x\}=\{2017+2017 r\}=\{2017 r\}=\frac{1}{2017}$, so $2017 r=n+\frac{1}{2017}$, where $n$ is an integer. Dividing both sides of this equation by $2017$ yields $r=\frac{n}{2017}+\frac{1}{2017^{2}}$, where $n=0,1,2, \ldots, 2016$ so that we have $0 \leq r<1$. Thus, we have $x=1+r=1+\frac{n}{2017}+\frac{1}{2017^{2}}$ for $n=0,1,2, \ldots, 2016$. The sum of these solutions is $2017 \cdot 1+\frac{2016 \cdot 2017}{2} \cdot \frac{1}{2017}+2017 \cdot \frac{1}{2017^{2}}=2017+\frac{2016}{2}+\frac{1}{2017}=3025 \frac{1}{2017}$.