Sum of all natural numbers. Okay, I do know that there are three ways of showing that it equals $-1\over 12$: 
1) The Reimann zeta function calculated for $-1$(see picture)
2) The one involving Grandi's series, the series $1-2+3-4+5..$ , and then finally getting $1+2+3+4+5...$.
3)Ramanujan's method of equating it to a constant, and subtracting $4$ times that equation from itself, and solving.
But, I have also seen a person solve it this way:
$$1+2+3+4+5+6..=n$$
$$1+(2+3+4)+(5+6+7)+(8+9+10)..=n$$
$$1+9+18+27+...=n$$
$$1+9(1+2+3..)=n$$
$$1+9n=n$$
$$1=-8n$$
$$n=\frac{-1}{8}$$
I am torn between the two. I know that the previous value is used and accepted by string theorists, but what about the second one? Is it valid mathematically? Or are both, due to infinity being infinity?
Edit: I added this picture after YiFan's answer.

 A: A series: $$\sum_{k=1}^\infty {u_k} $$
may converge only if $$\lim_{k\to\infty} u_k = 0$$
The ratio test proves that the natural numbers fail this property completely: Note that $$\frac {N+1}{N}=1+\frac 1N > 1 \space \forall N \in \Bbb N $$
In other words, the series you suggest is divergent.
A: None of these values are correct. The only remotely "right" one I'd probably due to the Riemann zeta function. However, it's not as easy as many people think. See, the Riemann zeta function is only defined as the sum $1^{-s}+2^{-s}+3^{-s}+...$ when $\operatorname{Re}s>1$. When we talk about the value of the function at values outside this domain, we are really talking about the analytic continuation of this function, defined as an integral in terms of the gamma function. See, for example, the wikipedia page on the Riemann zeta function. The reason why string theorists might find the value $-1/12$ useful is due to this property (of course I can't know for sure, I'm not a string theorist). I do know for sure, however, that it's not due to some nonsense like $1+2+...=-1/12$.
So, both the values $-1/8$ and $-1/12$ are equally wrong. It's about time this misconception is dispelled.

Despite the above remarks, I think it's useful to explain exactly why the reasoning in your question by grouping doesn't work.
Basically, the problem is that you cannot simply group terms together in a divergent series, or otherwise rearrange them, because you will end up changing the value of the series. A more basic example is the sum $1-1+1-1+...$. Is it equal to $1+(-1+1)+(-1+1)+...=1$ or $(1-1)+(1-1)+...=0$? (The unfortunate numberphile video promotes the misconception that this sum is in fact $1/2$ because it is "halfway in between $1$ and $0$, which is a purely ridiculous argument.) Rearrangements and groupings of terms is however permitted when your series converges absolutely, which is a neat theorem taught in any introductory analysis class.
