Find all triples $(x,y,z)$ of real numbers that satisfy the system
$\begin{cases} x + y + z = 2008 \\ x^2 + y^2 + z^2 = 6024^2 \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2008} \end{cases}$






To find all triples \((x, y, z)\) of real numbers that satisfy the given system of equations:

\[
\begin{cases}
x + y + z = 2008 \\
x^2 + y^2 + z^2 = 6024^2 \\
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2008}
\end{cases}
\]

we will proceed step-by-step.

1. **Sum of the variables:**
   \[
   x + y + z = 2008
   \]

2. **Sum of the squares of the variables:**
   \[
   x^2 + y^2 + z^2 = 6024^2
   \]

3. **Sum of the reciprocals of the variables:**
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2008}
   \]

   Using the identity for the sum of reciprocals:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{xy + yz + zx}{xyz}
   \]
   we can rewrite the third equation as:
   \[
   \frac{xy + yz + zx}{xyz} = \frac{1}{2008}
   \]

4. **Expressing the product \(xyz\):**
   From the third equation, we get:
   \[
   xy + yz + zx = \frac{xyz}{2008}
   \]

5. **Using the identity for the sum of squares:**
   We know:
   \[
   (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
   \]
   Substituting the known values:
   \[
   2008^2 = 6024^2 + 2(xy + yz + zx)
   \]
   Simplifying:
   \[
   2008^2 = 6024^2 + 2(xy + yz + zx)
   \]
   \[
   2008^2 - 6024^2 = 2(xy + yz + zx)
   \]
   \[
   2(xy + yz + zx) = 2008^2 - 6024^2
   \]
   \[
   xy + yz + zx = \frac{2008^2 - 6024^2}{2}
   \]

6. **Calculating \(2008^2 - 6024^2\):**
   \[
   2008^2 = 2008 \times 2008 = 4032064
   \]
   \[
   6024^2 = 6024 \times 6024 = 36288876
   \]
   \[
   2008^2 - 6024^2 = 4032064 - 36288876 = -32256812
   \]
   \[
   xy + yz + zx = \frac{-32256812}{2} = -16128406
   \]

7. **Finding the product \(xyz\):**
   From the third equation:
   \[
   xy + yz + zx = \frac{xyz}{2008}
   \]
   Substituting the value of \(xy + yz + zx\):
   \[
   -16128406 = \frac{xyz}{2008}
   \]
   \[
   xyz = -16128406 \times 2008 = -32356816 \times 2008
   \]

8. **Forming the polynomial:**
   The polynomial whose roots are \(x, y, z\) is:
   \[
   t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0
   \]
   Substituting the known values:
   \[
   t^3 - 2008t^2 - 16128406t + 32356816 \times 2008 = 0
   \]

9. **Simplifying the polynomial:**
   Let \(a = 2008\), then the polynomial becomes:
   \[
   t^3 - at^2 - 4a^2t + 4a^3 = 0
   \]
   Substituting \(a = 2008\):
   \[
   t^3 - 2008t^2 - 4 \cdot 2008^2 t + 4 \cdot 2008^3 = 0
   \]

10. **Finding the roots:**
    By inspection or using polynomial root-finding techniques, we find the roots to be:
    \[
    t = 2008, 4016, -4016
    \]

Thus, the solutions are:
\[
(x, y, z) = (2008, 4016, -4016)
\]

The final answer is \( \boxed{ (2008, 4016, -4016) } \).