Let $V=[(x,y,z)|0\le x,y,z\le 2008]$ be a set of points in a 3-D space.
If the distance between two points is either $1, \sqrt{2}, 2$, we color the two points differently.
How many colors are needed to color all points in $V$?
1. **Claim**: We need 7 colors to color all points in \( V \).

2. **Example of 7 Points**: Consider the seven points \((2,2,2)\), \((3,2,2)\), \((1,2,2)\), \((2,3,2)\), \((2,1,2)\), \((2,2,3)\), and \((2,2,1)\). These points are such that the distance between any two of them is either \(1\), \(\sqrt{2}\), or \(2\). Therefore, each of these points must be colored differently.

3. **Coloring Construction**: Denote the seven colors as \( C_0, C_1, C_2, C_3, C_4, C_5, C_6 \). We will use a periodic coloring scheme where \( C_{7k+l} = C_l \).

4. **Color Assignment**: Assign the color of the point \((x, y, z)\) to be \( C_{x + 2y + 3z} \).

5. **Function Definition**: Define \( f(x, y, z) = x + 2y + 3z \). The colors of two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the same if and only if \( f(x_1, y_1, z_1) = f(x_2, y_2, z_2) \).

6. **Case Analysis**:
    - **Case 1: Distance is 1**. If the distance between two points is 1, then two of the coordinates are the same and the third differs by 1. Therefore, \( f(P_1) \neq f(P_2) \).
    - **Case 2: Distance is \(\sqrt{2}\)**. If the distance between two points is \(\sqrt{2}\), then one coordinate is the same and the other two differ by 1. Thus, \( f(P_1) \neq f(P_2) \).
    - **Case 3: Distance is 2**. If the distance between two points is 2, then two coordinates are the same and the third differs by 2. Hence, \( f(P_1) \neq f(P_2) \).

7. **Conclusion**: Since the function \( f(x, y, z) = x + 2y + 3z \) ensures that no two points at the specified distances have the same color, the coloring scheme works with 7 colors.

\(\blacksquare\)

The final answer is \( \boxed{7} \)