, 2020

Solution:
For irrational $\frac{1}{\sqrt{r}}, -\frac{1}{\sqrt{r}}$ must also be a root of $P$. Therefore
$$P(x) = \frac{\left(x^{2} - \frac{1}{2}\right)\left(x^{2} - \frac{1}{3}\right) \cdots \left(x^{2} - \frac{1}{1000}\right)}{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right) \cdots \left(x + \frac{1}{31}\right)}$$
We get the sum of the coefficients of $P$ by setting $x=1$, so we use telescoping to get
$$P(1) = \frac{\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{999}{1000}}{\frac{3}{2} \cdot \frac{4}{3} \cdots \frac{32}{31}} = \frac{1}{16000}$$