Answer: 200

Solution:

Group the equation as $$\begin{aligned} &2(1000x^6 - 1) \\ &\quad {}+ x(100x^4 + 10x^2 + 1) = 0. \end{aligned}$$ Since $1000x^6 - 1 = (10x^2)^3 - 1$ $= (10x^2 - 1)$ $(100x^4 + 10x^2 + 1),$ the left side factors as $$\begin{aligned} &(100x^4 + 10x^2 + 1) \\ &\quad \big(2(10x^2 - 1) + x\big) \\ &= (100x^4 + 10x^2 + 1) \\ &\quad (20x^2 + x - 2). \end{aligned}$$

The quartic factor is always positive, so the two real roots are the roots of $20x^2 + x - 2 = 0,$ namely $x = \frac{-1 \pm \sqrt{161}}{40}.$ The root of the form $\frac{m + \sqrt{n}}{r}$ is $\frac{-1 + \sqrt{161}}{40},$ with $m = -1,$ $n = 161,$ $r = 40,$ and $\gcd(-1, 40) = 1.$ Thus $m + n + r = -1 + 161 + 40$ $= 200.$