All right, well, here's the answer. It might seem a little "out-there" for the first-time reader (I would say it would even look almost magical), but it's not that imaginative, once you're used to these manipulations:
Starting from the geometric sum \(S(x)\):
\[\begin{align}
S(x)&=1+x+x^2+\cdots\implies\\
xS(x)&=x+x^2+x^3+\cdots\implies\\
1+xS(x)&=1+x+x^2+\cdots=S(x)\implies\\
S(x)&=\frac{1}{1-x}=1+x+x^2+\cdots
\end{align}
\]All right, so, we take the derivative (this is one of those apparently 'magical' parts, which becomes pretty mundane after a while). If you've studied generating functions, chances are this is nothing new:
\[
S'(x)=\frac{1}{(x-1)^2}=0+1+2x+3x^2\cdots
\]But, we can't get a definite sum for this series, since \(S'(x)\) has a pole at \(x=1\). So, with some foresight, we plug in \(x=-1\) (it will become obvious as of why, later):
\[
S'(-1)=\frac{1}{4}=1-2+3-4+\cdots
\](This was the second hint I gave, about Abel-summability. This shouldn't have to be derived, I'm just doing it as a little extra).
All right, so, we now use the function \(\zeta(s)\) as a manipulation aid:
\[
\zeta(s)=\sum_{n=1}^\infty n^{-s}
\]We wish to negate every even term of the series, to be able to solve back for our own, final value, so we begin:
\[
\begin{align}
\zeta(s)&=1^{-s}+2^{-s}+3^{-s}+\cdots\implies\\
(1-2^{-s})\zeta(s)&=1^{-s}+2^{-s}-2^{-s}+3^{-s}+\cdots=1^{-s}+0+3^{-s}+\cdots\implies\\
(1-4^{-s})\zeta(s)&=1^{-s}-2^{-s}+3^{-s}+\cdots
\end{align}
\]Which is what we wished to find. Now, we know the value of this last series manipulation at \(s=-1\), so we use this to find our value for \(\zeta(-1)\), which is the sum of the natural set:
\[\begin{align}
(1-4^{1})\zeta(-1)&=1-2+3-4+\cdots=\frac{1}{4}\\
(-3)\zeta(-1)&=\frac{1}{4}\\
\zeta(-1)&=-\frac{1}{12}
\end{align}
\]So... guess what? The sum of all natural number is \(-\frac{1}{12}\).
Enjoy!
http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7_%C2%B7_%C2%B7