How many positive roots does polynomial $x^{2002} + a_{2001}x^{2001} + a_{2000}x^{2000} + \cdots + a_1x + a_0$ have such that $a_{2001} = 2002$ and $a_k = -k - 1$ for $0\leq k \leq 2000$?

$ 
\textbf{a)}\ 0
\qquad\textbf{b)}\ 1
\qquad\textbf{c)}\ 2
\qquad\textbf{d)}\ 1001
\qquad\textbf{e)}\ 2002
$
To determine the number of positive roots of the polynomial \( P(x) = x^{2002} + 2002x^{2001} - 2001x^{2000} - 2000x^{1999} - \cdots - 2x - 1 \), we can use Descartes' Rule of Signs. Descartes' Rule of Signs states that the number of positive real roots of a polynomial is either equal to the number of sign changes between consecutive non-zero coefficients or less than it by an even number.

1. **Identify the coefficients and their signs:**
   \[
   P(x) = x^{2002} + 2002x^{2001} - 2001x^{2000} - 2000x^{1999} - \cdots - 2x - 1
   \]
   The coefficients are:
   \[
   1, 2002, -2001, -2000, \ldots, -2, -1
   \]

2. **Count the sign changes:**
   - From \(1\) to \(2002\): no sign change.
   - From \(2002\) to \(-2001\): one sign change.
   - From \(-2001\) to \(-2000\): no sign change.
   - From \(-2000\) to \(-1999\): no sign change.
   - Continue this pattern until the last coefficient \(-1\).

   We observe that there is only one sign change from \(2002\) to \(-2001\).

3. **Apply Descartes' Rule of Signs:**
   According to Descartes' Rule of Signs, the number of positive roots is equal to the number of sign changes or less by an even number. Since there is only one sign change, the polynomial \( P(x) \) has exactly one positive root.

Conclusion:
\[
\boxed{1}
\]