jmc

Because the problem asks for the sum of the lengths of the intervals, we can replace $x$ with $x-2010,$ and the answer will not change. Then we have the inequality $$\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge 1.$$Let $f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}$. Note that $f(x)$ has three vertical asymptotes at $x=-1,0,1.$ Since the function $g(x)=1/x$ is decreasing over every connected interval on which it is defined, the same is true of $f(x).$ That is, $f(x)$ is decreasing on each of the intervals $(-\infty ,-1),$ $(-1,0),$ $(0,1),$ and $(1,\infty ).$

As $x$ approaches $\infty$ and as $x$ approaches $-\infty ,$ we see that $f(x)$ approaches $0.$ And as $x$ approaches each of the vertical asymptotes $x=-1,0,1$ from the left, $f(x)$ approaches $-\infty ,$ while as $x$ approaches each of the vertical asymptotes from the right, $f(x)$ approaches $\infty .$ This lets us sketch the graph of $f(x),$ as shown below: Therefore, the equation $f(x)=1$ has three roots $p,q,r,$ where $p\in (-1,0),$ $q\in (0,1),$ and $r\in (1,\infty ).$ In terms of these roots, the values of $x$ such that $f(x)\ge 1$ are $$(-1,p]\cup (0,q]\cup (1,r].$$The sum of the lengths of the three intervals above is $(p+1)+q+(r-1)=p+q+r,$ so we want to find the sum of the roots of $f(x)=1.$

Multiplying the equation $f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1$ by $(x+1)x(x-1)$ and then rearranging, we get the cubic equation $$x^3-3x^2-x+1=0.$$By Vieta's formulas, the sum of the roots of this equation is $\boxed{3}.$