$\sum _{r=1}^{100}(\frac{1}{99+r}-\frac{1}{2r(2r-1)})=$

$\sum _{r=1}^{100}\frac{1}{2r(2r-1)}=\sum _{r=1}^{100}(\frac{1}{2r-1}-\frac{1}{2r})$

$=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\mathrm{........}+\frac{1}{199}-\frac{1}{200}\right)$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\mathrm{......}+\frac{1}{200})-2(\frac{1}{2}+\frac{1}{4}+\mathrm{......}+\frac{1}{200})$

$\therefore E=\sum _{r=1}^{100}\frac{1}{99+r}-\sum _{r=101}^{200}\frac{1}{r}$

$=\frac{1}{100}-\frac{1}{200}=\frac{1}{200}$