Basis step: For $n=1$, equation holds.
Inductive step: Now, $$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i} \tag{i. h.}$$ Now we want to show that $$\sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^{n+1}_{i=1}\frac{1}{n+1+i}.$$ So, $$ \sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i} = \sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i} + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2} = \text{(by inductive hypothesis)}= \sum^n_{i=1}\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2} $$ This is where I'm stuck.
Change the indices in the sum: $$\sum_{i=1}^n\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{i=0}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\ =\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{i=1}^{n+1}\frac{1}{n+i+1}.$$
A straightforward approach is as follows.
$$\begin{align} \sum_{i=1}^{2n}\frac{(-1)^{i-1}}{i}&=\sum_{i=1}^n\left(\frac{1}{2i-1}-\frac{1}{2i}\right)\\\\ &=\sum_{i=1}^n\left(\frac{1}{2i-1}+\frac{1}{2i}\right)-\sum_{i=1}^n\frac1i\\\\ &=\sum_{i=1}^{2n}\frac1i-\sum_{i=1}^n\frac1i\\\\ &=\sum_{i=n+1}^{2n}\frac1i\\\\ &=\sum_{i=1}^n\frac{1}{n+i} \end{align}$$
