$$
\left\{\begin{array}{l}
2006(x-y)+2000(y-z)+2008(z-x)=0, \\
2000^{2}(x-y)+2000^{2}(y-z)+2000^{2}(z-x)=2008 .
\end{array}\right.
$$

Then the value of $z-y$ is $\qquad$ .
$=、 1.2008$
Let $z-y=t$.
From the first equation, we get $z-2 x+y=0$.
Thus, $z-x=\frac{t}{2}$, and consequently, $x-y=\frac{t}{2}$.
Therefore, from the second equation, we can find that
$$
t=z-y=2008 .
$$