Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum
\[f \left(\frac{1}{2019} \right)-f  \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]
$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$
First, note that $f(x) = f(1-x)$. We can see this since 
\[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\]
Using this result, we regroup the terms accordingly:
\[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\]
\[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\]
Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{\textbf{(A) }0}$.