A function with the reciprocal values with the roots of $p(x)$ is $p(\frac{1}{x})$ $p(\frac{1}{x})=(\frac{1}{x})^{1999}+2(\frac{1}{x})^{1998}+3(\frac{1}{x})^{1997}+\ldots+2000 = 0$ $(\frac{1}{x})^{1999}+2(\frac{1}{x})^{1998}+3(\frac{1}{x})^{1997}+\ldots+2000 = 0$ $1+2x+3x^2+\ldots+2000x^{1999} = 0$ So a polynomial with the reciprocal values of the roots of $p(x)$ is $1+2x+3x^2+\ldots+2000x^{1999}$