Solution:

The answer is (C). We first observe that the numbers $2001$ and $2003$ are not possible, since they are odd numbers. Indeed, since $P(2000) = a_{n} 2000^{n} + a_{n-1} 2000^{n-1} + \cdots + a_{1} 2000 + a_{0} = 2000$ is an even number, $a_{0}$ must be an even number, and consequently, computing the value of the polynomial at any even number $2k$, we get that $P(2k) = a_{n} (2k)^{n} + a_{n-1} (2k)^{n-1} + \cdots + a_{1} (2k) + a_{0}$ must be even.

Conversely, it is easy to construct examples in which the other values can be obtained:
$$$$
$$\begin{aligned} \text{if } P(X) = 2001 - (X - 2001)^{2}, & \quad P(2002) = 2000 ; \\ \text{if } P(X) = X, & \quad P(2002) = 2002 ; \\ \text{if } P(X) = (X - 2000)^{2} + 2000, & \quad P(2002) = 2004 . \end{aligned}$$