$ab=1991(a+b)$ for $a=\lfloor x\rfloor$ and $b=\{x\}$. Then $(a-1991)(b-1991)=1991^2$. Then $b=\frac{1991a}{a-1991}=1991+\frac{1991^2}{a-1991}$ is between $0$ and $1$. Then $\frac{1991^2}{a-1991}$ is between $-1991$ and $-1990$ so $a-1991$ is between $\frac{1991^2}{-1992}=-1992.0005$ and $-1991$ so $a$ is between $-1.0005$ and $0$. Hence $a=0$ or $-1$, and so $b=0$ or $\frac{1991}{1992}$. Hence $x=0$ or $x=-\frac{1}{1992}$.