Let $f$ be a real-valued function such that
\[f(x) + 2f(\frac{2002}{x}) = 3x\]
for all $x>0.$ Find $f(2).$
$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$
When $x = 2$, then we get $f(2) + 2f(1001) = 6$; we can also substitute $x$ as $1001$, then we will get $f(1001) + 2f(2) =3003$. Solve this system of equations, then we get $f(2)= 2000$ $\Longrightarrow \boxed{\mathrm{B}}$.