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 $0$?

Almost everything in your proof works fine until you write "this looks ok if I interprete it as..." 
Until then, you are manipulating infinite sums of series with positive terms, these are extended nonnegative real numbers (numbers $x$ such that $0\leqslant x\leqslant+\infty$, if you like) hence adding them and equating them is perfectly legal.
The trouble begins when you substract them, since there is no substraction on the set of extended nonnegative real numbers. Unsurprisingly, you soon must deal with $(+\infty)-(+\infty)$ differences, and chaos ensues.
A less sophisticated example, flawed quite similarly, is to start with the correct identity
$$
1+1+1+\cdots=\underline{\mathbf 1}+(\color{red}{1}+\color{blue}{1}+\color{green}{1}+\cdots),
$$
and to deduce from it that
$$
0=(1-\color{red}{1})+(1-\color{blue}{1})+(1-\color{green}{1})+\cdots=\underline{\mathbf 1}.
$$
