We need to determine whether there exist $2002$ distinct positive integers $k_1, k_2, \ldots, k_{2002}$ such that for any positive integer $n \geq 2001$, at least one of $k_1 2^n + 1, k_2 2^n + 1, \ldots, k_{2002} 2^n + 1$ is prime.

To address this, we generalize the problem for $F > 2002$. Consider choosing $X = F^F \prod_{i=1}^{F^F} (p_i - 1)$, where $p_i$ are distinct primes. This choice ensures that $X$ is extremely large.

By Fermat's Little Theorem, for any prime $p_i$ dividing $k_x + 1$, we have:
$$2^X k_i + 1 \equiv k_x + 1 \equiv 0 \pmod{p_i}.$$

Given the size of $2^X k_i + 1$ being greater than $p_i$, it follows that $2^X k_i + 1$ is not prime. Thus, no such integers $k_1, k_2, \ldots, k_{2002}$ exist.

Therefore, the answer is: \boxed{\text{No}}.