Let $f$ be a real-valued function such that \[f(x)+2f\left(\dfrac{2002}x\right)=3x\] for all $x>0$.  Find $f(2)$.

$\textbf{(A) }1000\qquad\textbf{(B) }2000\qquad\textbf{(C) }3000\qquad\textbf{(D) }4000\qquad\textbf{(E) }6000$
1. Given the functional equation:
   \[
   f(x) + 2f\left(\dfrac{2002}{x}\right) = 3x \quad \text{for all } x > 0
   \]
   We need to find \( f(2) \).

2. Substitute \( \frac{2002}{x} \) for \( x \) in the given equation:
   \[
   f\left(\frac{2002}{x}\right) + 2f(x) = 3 \left(\frac{2002}{x}\right)
   \]
   Simplifying the right-hand side:
   \[
   f\left(\frac{2002}{x}\right) + 2f(x) = \frac{6006}{x}
   \]

3. Now we have a system of equations:
   \[
   \begin{cases}
   f(x) + 2f\left(\dfrac{2002}{x}\right) = 3x \\
   f\left(\dfrac{2002}{x}\right) + 2f(x) = \dfrac{6006}{x}
   \end{cases}
   \]

4. Let \( f(x) = a \) and \( f\left(\dfrac{2002}{x}\right) = b \). Then the system becomes:
   \[
   \begin{cases}
   a + 2b = 3x \\
   b + 2a = \dfrac{6006}{x}
   \end{cases}
   \]

5. To solve for \( a \) and \( b \), we add the two equations:
   \[
   (a + 2b) + (b + 2a) = 3x + \dfrac{6006}{x}
   \]
   Simplifying, we get:
   \[
   3a + 3b = 3x + \dfrac{6006}{x}
   \]
   Dividing by 3:
   \[
   a + b = x + \dfrac{2002}{x}
   \]

6. Now, we solve for \( a \) (which is \( f(x) \)):
   \[
   a = x + \dfrac{2002}{x} - b
   \]

7. Substitute \( b \) from the first equation \( a + 2b = 3x \):
   \[
   b = \dfrac{3x - a}{2}
   \]

8. Substitute \( b \) into \( a + b = x + \dfrac{2002}{x} \):
   \[
   a + \dfrac{3x - a}{2} = x + \dfrac{2002}{x}
   \]
   Multiply through by 2 to clear the fraction:
   \[
   2a + (3x - a) = 2x + \dfrac{4004}{x}
   \]
   Simplify:
   \[
   a + 3x = 2x + \dfrac{4004}{x}
   \]
   Solving for \( a \):
   \[
   a = \dfrac{4004}{x} - x
   \]

9. Therefore, the function \( f(x) \) is:
   \[
   f(x) = \dfrac{4004}{x} - x
   \]

10. To find \( f(2) \):
    \[
    f(2) = \dfrac{4004}{2} - 2 = 2002 - 2 = 2000
    \]

The final answer is \( \boxed{2000} \).