Find the number of ordered pairs of real numbers $(a,b)$ such that $(a + bi)^{2002} = a - bi$.
Let $z = a + bi$, so $\overline{z}= a - bi$. Then the given relation becomes $z^{2002} = \overline{z}$. Note that
$$|z|^{2002} = \left|z^{2002}\right| = |\overline{z}| = |z|,$$from which it follows that
$$|z|\left(|z|^{2001} - 1\right) = 0.$$Hence, $|z| = 0$ or $|z| = 1$.

If $|z| = 0,$ then $z = 0,$ and $(a,b) = (0,0).$

If $|z|=1$, then we have $z^{2002} = \overline{z}$, which is equivalent to $z^{2003} = \overline{z}\cdot z = |z|^2 = 1$.  The equation $z^{2003} = 1$ has $2003$ distinct solutions, which gives us 2003 pairs $(a,b)$.

Therefore, there are altogether $1 + 2003 = \boxed{2004}$ ordered pairs that meet the required conditions.