There are so many

$\begin{aligned} \dfrac{20102011}{2012} & = 9991 + \frac{119}{2012} = 9991+\frac{1}{\frac{2012}{119}} \\ & = 9991+\frac{1}{16 + \frac{108}{119}} = 9991+\frac{1}{16+\frac{1}{\frac{119}{108}}} \\ & = 9991+\frac{1}{16+\frac{1}{1 + \frac{11}{108}}} = 9991+\frac{1}{16+\frac{1}{1+\frac{1}{\frac{108}{11}}}} \\ & = 9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{9}{11}}}} = 9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{\frac{11}{9}}}}} \\ & = 9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{1+\frac{2}{9}}}}} = 9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{1+\frac{1}{\frac{9}{2}}}}}} \\ & = 9991+\dfrac{1}{16+\dfrac{1}{1+\dfrac{1}{9+\dfrac{1}{1+\dfrac{1}{4 + \dfrac{1}{2}}}}}} \end{aligned}$

$\Rightarrow a-b+c-d+e-f+g = 9991-16+1-9+1-4+2 = \boxed{9966}$

Read up on continued fractions to solve this using Mr. Cheong's method.

Python 3.4.2

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def convergents(num,den):
    a=[]
    if den==1:
        a.append(num)
        return a
    else:
        a.append(num//den)
        return a+convergents(den,num%den)
print(convergents(20102011,2012)) #Returns the list [9991, 16, 1, 9, 1, 4, 2]
9991-16+1-9+1-4+2 #Returns 9966

PS: This is a function from my library which is provided 'as-is'. Thus, I had to calculate $a-b+c-d+e-f+g$ manually.