jmc

By Vieta's formulas, $$\{\begin{array}{rl}a+b+c& =0\\ ab+bc+ac& =-2011.\end{array}$$Since $a+b=-c,$ the second equation becomes $ab+(-c)c=-2011$, or $$c^2-ab=2011.$$At least two of $a,b,c$ must have the same sign; without loss of generality, let $a$ and $b$ have the same sign. Furthermore, since we can negate all of $a,b,c$ and still satisfy the two above equations, assume that $c\ge 0.$ (Note that we only want the sum $|a|+|b|+|c|$, which does not change if we swap or negate the variables.)

Now, we have $ab\ge 0,$ so $c^2\ge 2011$, giving $c\ge 44.$ We also have $$\frac{c^2}{4}={\left(\frac{a+b}{2}\right)}^2\ge ab$$by AM-GM, so $2011=c^2-ab\ge 3c^2/4,$ giving $c\le 51.$

Finally, we have $(a-b{)}^2=(a+b{)}^2-4ab=(-c{)}^2-4(c^2-2011)=8044-3c^2$, which must be a perfect square.

Testing $c=44,45,\dots ,51$, we find that $8044-3c^2$ is a perfect square only when $c=49$. Therefore, $c=49$, and so $$\{\begin{array}{rl}a+b& =-c=-49,\\ ab& =c^2-2011=390.\end{array}$$Thus, $a$ and $b$ are the roots of $t^2+49t+390=0$, which factors as $(t+10)(t+39)=0$. Thus, $\{a,b\}=\{-10,-39\}$.

The answer is $$|a|+|b|+|c|=39+10+49=\boxed{98}.$$