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 $0$?
 A: 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
$$
0=(1-\color{red}{1})+(1-\color{blue}{1})+(1-\color{green}{1})+\cdots=\underline{\mathbf 1}.
$$
A: 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.
A: Your proof is not right since $x=\infty$.
A: You are manipulating divergent series. When you do the substraction, you can in fact obtain any limit you want by changing the order in which you are adding or subtracting things.
For example, 
$(1 - \frac 12 - \frac 14 - \frac 18 - \ldots) + (\frac 13 - \frac 16 - \frac 1{12} - \frac 1{24}- \ldots) + (\frac 15 - \frac 1{10} - \frac 1{20} - \frac 1{40} - \ldots) + \ldots = 0 + 0 + 0 + \ldots = 0$
