$x$ and the biggest integer $\leq x$

$\begin{aligned} 2016x & = 2017\lfloor x \rfloor & \small \color{#3D99F6} \text{Note that }x = \lfloor x \rfloor + \{x\} \\ 2016(\lfloor x \rfloor + \{x\}) & = 2017\lfloor x \rfloor & \small \color{#3D99F6} \text{where }\{x\} \text{ is the fractional part of }x \\ \implies 2016 \{x\} & = \lfloor x \rfloor \end{aligned}$

Note that the $LHS = 2016 \{x\} > 0$ for all $x$ , while the $RHS = \lfloor x \rfloor < 0$ for $x < 0$ . Therefore, there is no solution for $x <0$ .

Note also that $RHS = \lfloor x \rfloor$ is integral and for $LHS = 2016 \{x\}$ to be integral the solution is $\{x\} = \dfrac k{2016}$ , where $k = 0, 1, 2 ... 2015$ . Therefore, there are $\boxed{2016}$ solutions.

Note: The solution is $x = \lfloor x \rfloor + \dfrac {\lfloor x \rfloor}{2016}$ for $\lfloor x \rfloor \in [0, 2015]$ .