1. Given the product of four real numbers is 1, and the sum of any one of the numbers with the product of the other three is 1000. Then the sum of these four numbers is
$=、 1.2000$. 
These four real numbers are all non-zero and not all equal, denoted as $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$. Taking any one of these numbers $x_{i} (i=1,2,3,4)$, the other three numbers are $x_{j}$, $x_{k}$, $x_{r}$, then we have
$$
\left\{\begin{array}{l}
x_{i}+x_{j} x_{k} x_{r}=1000, \\
x_{i} x_{j} x_{k} x_{r}=1 .
\end{array}\right.
$$
Since $x_{i}$ is any one of the four numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$,
Therefore, $x_{i}$, $x_{j}$, $x_{k}$, $x_{r}$ are all roots of the equation $x^{2}-1000 x+1=0$.
Since the four numbers are not all equal, without loss of generality, let $x_{i} > x_{j}$, we have
$$
\begin{array}{r}
x_{i}=\frac{1000+\sqrt{1000^{2}-4}}{2}, \\
x_{j}=\frac{1000-\sqrt{1000^{2}-4}}{2} .
\end{array}
$$

Similarly, $x_{4}=\frac{1000+\sqrt{1000^{2}-4}}{2}$,
$$
x_{r}=\frac{1000-\sqrt{1000^{2}-4}}{2} \text {. }
$$

Therefore, the sum of the four real numbers is 2000.