How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$? 
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?

Let
$$\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\
y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\ 
z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}$$ 
so we have
$$x = y + z.$$
However, $x = 2\cdot z$, so $y$ = $z$ or
$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots  = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$ 
This looks ok if I interpret it as
$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} -  \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$ 
However, it's a bit weird if I write it as
$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} -  \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots  = 0.$$ 
How can a sum of positive numbers equal $0$?

Answering only "How can a sum of only positive numbers equal 0?" : look at $1^2 + 2^2 + 3^2 + 4^2 + 5^2+ ... = 0$ which is a meaningful result if the series is understood as zeta-series.
