For any real number $x$, we denote $\lfloor x\rfloor$ as the integer part of $x$, that is, the greatest integer less than or equal to $x$. We denote $\{x\}$ as its decimal part, that is, $\{x\}=x-\lfloor x\rfloor$. For example, we have $\lfloor 3.1\rfloor=3$ and $\{3.1\}=3.1-3=0.1$. We also have $\lfloor-2.7\rfloor=-3$ and $\{-2.7\}=-2.7-(-3)=0.3$.

Find all real numbers $x$ such that $\lfloor x\rfloor \times\{x\}=2019 \times x$.
Let $x$ be a solution to the problem. We denote $n=\lfloor x\rfloor$ and $y=\{x\}$. Then $n$ is an integer and $0 \leqslant y < 1$. We have $n+y=x$ and $n-2019y=n-2019\{x\}=n-2019(x-n)=2019n-x$. 

If $n>0$, then $2019 n>n$, so $n-2019>0$, thus $y>1$, which is impossible. If $0<n<2019$, then $0>n-2019$, so $y<0$, which is also impossible. Therefore, we must have $n \leqslant 0$, so we set $n=-n'$ with $n' \geqslant 0$. Then we have $y=\frac{-2019 n'}{-n'-2019}=\frac{2019 n'}{n'+2019}$.

But we also have $y<1$, so $2019 n'<n'+2019$, which simplifies to $2018 n'<2019$, thus $n'<\frac{2019}{2018}$, so $n'$ is 0 or 1, hence $n$ is 0 or -1. In the first case, we have $y=0$ so $x=0$. In the second case, we have $y=\frac{2019}{2020}$, so $x=-1+\frac{2019}{2020}=-\frac{1}{2020}$. Therefore, there are two solutions, which are 0 and $-\frac{1}{2020}$.