jmc

Let $p(x)=x^{10}+(13x-1{)}^{10}.$ If $r$ is a root of $p(x),$ then $r^{10}+(13x-1{)}^{10}=0.$ Then $(13r-1{)}^{10}=-r^{10},$ so $$-1={\left(\frac{13r-1}{r}\right)}^{10}={\left(\frac{1}{r}-13\right)}^{10}.$$Then $\frac{1}{r}-13$ has magnitude 1, so $$\left(\frac{1}{r}-13\right)\left(\frac{1}{\overline{r}}-13\right)=1,$$so $$\left(\frac{1}{r_1}-13\right)\left(\frac{1}{{\overline{r}}_1}-13\right)+\cdots +\left(\frac{1}{r_5}-13\right)\left(\frac{1}{{\overline{r}}_5}-13\right)=5.$$Expanding, we get $$\frac{1}{r_1{\overline{r}}_1}+\cdots +\frac{1}{r_5{\overline{r}}_5}-13\left(\frac{1}{r_1}+\frac{1}{{\overline{r}}_1}+\cdots +\frac{1}{r_5}+\frac{1}{{\overline{r}}_5}\right)+5\cdot 169=5.$$We see that $\frac{1}{r_1},$ $\frac{1}{{\overline{r}}_1},$ $\dots ,$ $\frac{1}{r_5},$ $\frac{1}{{\overline{r}}_5}$ are the solutions to $${\left(\frac{1}{x}\right)}^{10}+{\left(\frac{13}{x}-1\right)}^{10}=0,$$or $1+(13-x{)}^{10}=0.$ The first few terms in the expansion as $$x^{10}-130x^9+\cdots =0,$$so by Vieta's formulas, $$\frac{1}{r_1}+\frac{1}{{\overline{r}}_1}+\cdots +\frac{1}{r_5}+\frac{1}{{\overline{r}}_5}=130.$$Hence, $$\frac{1}{r_1{\overline{r}}_1}+\cdots +\frac{1}{r_5{\overline{r}}_5}=13\cdot 130-5\cdot 169+5=\boxed{850}.$$