This looks very similar to problem A1 at IMO 1979. It has basically the same solution as well. Namely: $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{2019}-\frac{1}{2020}$ $=$ $1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{2019}+\frac{1}{2020}$ $-$ $2(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2018}+\frac{1}{2020})$ $=$ $\frac{1}{1011}+\frac{1}{1012}+\cdots +\frac{1}{2020}$ as desired.