smc

First, note that $f(x)=f(1-x)$. We can see this since $$f(x)=x^2(1-x{)}^2=(1-x{)}^2{x}^2=(1-x{)}^2{\left(1-\left(1-x\right)\right)}^2=f(1-x)$$ 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 $\boxed{0}$.