Floor & fractional equation

Left part of equation is integer always, then right part should be integer. Again, right part $\in \left( 0;2 \right)$ (Why don't equal $0$ ? If sum of two fractional part function equal $0$ , then both fractional part function should equal to $0$ , but $\left\{ \frac { \left\{ x \right\} }{ k } \right\} \neq 0$ , because $\left\{ x \right\} \neq 0$ ), then left part should $\in \left( 0;2 \right)$ . So, both part of equation should equal $1$ : $\left\lfloor \frac { \left\lfloor x \right\rfloor }{ k } \right\rfloor +\left\lfloor \frac { k }{ \left\lfloor x \right\rfloor } \right\rfloor =\left\{ \frac { k }{ \left\{ x \right\} } \right\} +\left\{ \frac { \left\{ x \right\} }{ k } \right\} \Leftrightarrow \begin{cases} \left\lfloor \frac { \left\lfloor x \right\rfloor }{ k } \right\rfloor +\left\lfloor \frac { k }{ \left\lfloor x \right\rfloor } \right\rfloor =1 \\ \left\{ \frac { k }{ \left\{ x \right\} } \right\} +\left\{ \frac { \left\{ x \right\} }{ k } \right\} =1 \end{cases}$ Second equation of system has roots for any $k\in\mathbb{Z}$ . Consider first equation of the system. Let's consider function below. $f\left( x \right) = \left| \left\lfloor \frac { k }{ \left\lfloor x \right\rfloor } \right\rfloor +\left\lfloor \frac { \left\lfloor x \right\rfloor }{ k } \right\rfloor \right|$ If we take $\left\lfloor x \right\rfloor =k\pm1$ (" $+$ ", if $k>0$ and " $-$ ", if $k<0$ ), then function reach minimum value. Convert function for $\left\lfloor x \right\rfloor =k\pm1$ : $\left| \left\lfloor \frac { k }{ k\pm 1 } \right\rfloor +\left\lfloor \frac { k\pm 1 }{ k } \right\rfloor \right| =\left| \left\lfloor 1\pm \frac { 1 }{ k } \right\rfloor \right|$

If $k\neq\pm 1$ then ${ f }_{ \min{} }\left( x \right)=1$ , aslo first equation of system has roots. If $k=\pm1$ then ${ f }_{ \min{} }\left( x \right)=2$ , aslo equation has no roots. So, answer is $\boxed{2}$ !