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: 
The most common way to define $\sum\_{n\geq1}n$ is by its partial sums:
$$
\sum\_{n\geq1}n\equiv\lim\_{N\rightarrow\infty}s\_{N}\text{ where }s\_{N}\equiv\sum\_{n=1}^{N}n=\frac{1}{2}N\left(N+1\right).
$$
Clearly, this is divergent.

---

As you seem to have noticed, there are other ways to define this sum. This causes all sorts of confusion, since people like to also use the notation $\sum\_{n\geq1}n$ in these alternative definitions.

Let's now look at "zeta function regularization", which is the interpretation which yields $-1/12$. In particular, note that
$$
\zeta(s)=\sum\_{n\geq1}n^{-s}\text{ for }\operatorname{Re}(s)>1
$$
by definition. In the above, we are using the usual notion of summation by a limit of partial sums. The $\zeta$ function, however, is also defined for other values of $s$ by [analytic continuation](https://en.wikipedia.org/wiki/Analytic_continuation). We can thus reinterpret
$$
\check{\sum\_{n\geq1}}n\equiv\zeta(-1)=-1/12.
$$
However, note that the two interpretations $$\sum\text{ and }\check{\sum}$$
are completely different!

---

As for the method which you refer to as "abusing divergent and oscillating series", perhaps you are thinking of [this](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Heuristics). This is not rigorous.
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:

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

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