Question: A fairly well-known (and perplexing) fact is that [the sum of all natural numbers is given the value -1/12](https://math.stackexchange.com/q/39802/31408), at least in certain contexts. Quite often, a ["proof"](https://www.youtube.com/watch?v=w-I6XTVZXww) is given which involves abusing divergent and oscillating series, and the numbers seem to fit at the end. (There are also other proofs involving the zeta function or other things, but that's not my concern right now).

However, I tried to calculate it my own way using similar methods, and this is what I got:

$$\begin{align}
S = 1 &+ 2 + 3 + 4 + …\\\\
2S = 1 &+ 2 + 3 + 4 + 5 + \ldots\\
 &+ 1 + 2 + 3 + 4 + \ldots\\\\
 = 1 &+ 3 + 5 + 7 + 9 + \ldots\\
\end{align}$$
Also,
$2S = 2 + 4 + 6 + 8 + 10 + \ldots$

Adding the above together,

$$4S = 1 + 2 + 3 + 4 + 5 + … = S$$

Which means $3S = 0$, therefore $S = 0$

Obviously, there must be some reason why this is not ok, otherwise we'd have $-1/12 = 0$.  

But why is my method wrong while the one involving oscillating series is considered acceptable?

Additional clarification: I was wondering if there are specific ways to manipulate this kind of series such that by assuming that the result is finite and performing only certain types of "permitted" operations, one could be confident to get to the same result as the rigorous way of assigning a finite value to the sum of the series. So far, the consensus seems to be negative.

Answer: 
In some sense, the error is in the subtraction. You're trying to do some arithmetic on things which are no longer proper numbers, and not everything behaves the same. 

For a mild simplification, think about the sum
$$ S = 1 + 1 + 1 + \cdots $$
Then by the same reasoning as yours, we can consider
$$ 2S := S + S = 1 + 1 + 1 + \cdots $$
Here I've interleaved sums, and this is okay because all terms are positive, though not valid in general. This addition of $S$ with itself in some sense is fine and we see $2S = S$. Then you'd like to say that
$$ 2S = S \implies S = 0, $$
but the problem it doesn't for arbitrary "quantities" S, including the case that $S = +\infty$, which is the value of $S$ here. (The above statement doesn't even hold in many finite algebraic systems like $\mathbb Z/6 \mathbb Z$.)

Weird things happen when you try to work with infinities, and not everything commutes. In particular, you need to be careful about subtracting infinite sums because [conditionally convergent sums can be rearranged to take arbitrary values](http://en.wikipedia.org/wiki/Riemann_series_theorem).
If you assume the sum of all naturals is finite and the "usual arithmetic" rules here, you can get pretty awesome results, like:

$$S=1+2+3+\ldots\implies 2+4+6+\ldots=2(1+2+3+\ldots)=2S\implies$$

$$S=1+2+3+\ldots=1+3+5+\ldots+2+4+6+\ldots=1+3+5+\ldots+2S\implies$$

$$-S=1+3+5+\ldots$$

And a lot more of whatever else you want, say:

$$1+1+1+\ldots-S=1+1+1+\ldots+1+3+5+\ldots=1+1+3+1+5+1+\ldots=$$

$$=2+4+6+\ldots=2S\implies 3S=1+1+1+\ldots\;,\;\;\text{etc.} $$