Consider the polynomial $p(x)=x^{1999}+2x^{1998}+3x^{1997}+\ldots+2000$. Find a nonzero polynomial whose roots are the reciprocal values of the roots of $p(x)$.
1. Given the polynomial \( p(x) = x^{1999} + 2x^{1998} + 3x^{1997} + \ldots + 2000 \), we need to find a polynomial whose roots are the reciprocals of the roots of \( p(x) \).

2. To find a polynomial whose roots are the reciprocals of the roots of \( p(x) \), we substitute \( x \) with \( \frac{1}{x} \) in \( p(x) \). This gives us:
   \[
   p\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^{1999} + 2\left(\frac{1}{x}\right)^{1998} + 3\left(\frac{1}{x}\right)^{1997} + \ldots + 2000
   \]

3. Simplify each term by expressing them with a common denominator:
   \[
   p\left(\frac{1}{x}\right) = \frac{1}{x^{1999}} + \frac{2}{x^{1998}} + \frac{3}{x^{1997}} + \ldots + 2000
   \]

4. Multiply through by \( x^{1999} \) to clear the denominators:
   \[
   x^{1999} \cdot p\left(\frac{1}{x}\right) = 1 + 2x + 3x^2 + \ldots + 2000x^{1999}
   \]

5. Therefore, the polynomial whose roots are the reciprocals of the roots of \( p(x) \) is:
   \[
   q(x) = 1 + 2x + 3x^2 + \ldots + 2000x^{1999}
   \]

Conclusion:
\[
\boxed{1 + 2x + 3x^2 + \ldots + 2000x^{1999}}
\]