1. Define the length of intervals $(m, n)$, $[m, n)$, $(m, n]$, and $[m, n]$ to be $n-m$ ($n, m \in \mathbf{R}$, and $n > m$). Then the sum of the lengths of the intervals of real numbers $x$ that satisfy
$$
\frac{1}{x-20}+\frac{1}{x-17} \geqslant \frac{1}{512}
$$

is $\qquad$
$-1.1024$.
Let $a=20, b=17, c=\frac{1}{512}$.
Then $a>b>c>0$.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant c$.
When $x>a$ or $x0, f(a)=b-a<0$.
Let the two real roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$. Then the interval of $x$ that satisfies $f(x) \leqslant 0$ is $\left(a, x_{2}\right]$, with the length of the interval being $x_{2}-a$.

Similarly, when $b<x<a$, the interval of $x$ that satisfies $f(x) \geqslant 0$ is $\left(b, x_{1}\right]$, with the length of the interval being $x_{1}-b$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{a c+b c+2}{c}=a+b+\frac{2}{c} \text {. }
$$

Then the sum of the lengths of the intervals of $x$ that satisfy the conditions is
$$
\begin{array}{l}
x_{2}-a+x_{1}-b=a+b+\frac{2}{c}-a-b \\
=\frac{2}{c}=1024 .
\end{array}
$$