Find all triples $(x, y, z)$ of real numbers that satisfy the system

$$
\left\{\begin{array}{l}
x+y+z=2008 \\
x^{2}+y^{2}+z^{2}=6024^{2} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2008}
\end{array}\right.
$$
The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.

$(x-2008)(y-2008)(z-2008)=0$.

Thus one of the variables equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-2008^{2}=2008^{2}(9-1)=$ $2 \cdot 4016^{2}$. Thus $(x, y, z)$ is the triple $(2008,4016,-4016)$ or any of its rearrangements.