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 $P(1)=\frac{\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{999}{1000}}{\frac{3}{2} \cdot \frac{4}{3} \cdots \frac{32}{31}}=\frac{1}{16000}$.