Floors Are Negligible

Relevant wiki: Floor and Ceiling Functions - Problem Solving

Let the fractional part of $x$ be $f$ where $0<f<1$ (as $x$ is non-integer) and the integer part be $y$ .

$x+\dfrac{2016}{x}=\lfloor x \rfloor +\dfrac{2016}{\lfloor x \rfloor} \implies y+f+\dfrac{2016}{y+f}=y+\dfrac{2016}{y}$

Multiplying through by $y(y+f)$ gives:

$\cancel{y^2 (y+f)} +y f(y+f)+\dfrac{2016 y \cancel{(y+f)}}{\cancel{y+f}}= \cancel{y^2 (y+f)} +\dfrac{2016 \cancel{y} (y+f)}{\cancel{y}} \implies y f(y+f)+2016y=2016y+2016f \implies y f(y+f)=2016f$

As $f \neq 0$ we have:

$y \cancel{f} (y+f)=2016 \cancel{f} \implies y(y+f)=2016 \implies f=\dfrac{2016-y^2}{y}$

We can now use $0<f<1$ to restrict the values of $y$ . First consider $y>0$ :

$0<\dfrac{2016-y^2}{y}<1 \implies 0<2016-y^2<y$

$\begin{cases} 2016-y^2 >0 & \implies y^2<2016 \implies y \leq 44 \\ 2016-y^2 < y & \implies 2016<y^2+y \implies y \geq 45 \end{cases}$

This shows there are no solutions for $y>0$ . Now we consider $y<0$ (so we have to flip the inequality when we multiply through by $y$ ):

$0<\dfrac{2016-y^2}{y}<1 \implies y<2016-y^2<0$

$\begin{cases} 2016-y^2>y & \implies 2016>y^2+y \implies y \geq -45 \\ 2016-y^2<0 & \implies y^2>2016 \implies y \leq -45 \end{cases}$

This shows us that $y=-45$ is the only solution for $y$ . Solving for $f$ gives:

$f=\dfrac{2016-y^2}{y} \implies f=\dfrac{2016-(-45)^2}{-45}=\dfrac{1}{5}$

This gives the value for $x$ as $y+f=-45+\dfrac{1}{5}=-44 \dfrac{4}{5}$ so the sum of the solutions is:

$\boxed{\boxed{-44 \dfrac{4}{5}}}$