Using the rules that prove the sum of all natural numbers is $-\frac{1}{12}$, how can you prove that the harmonic series diverges?

Letting $$ H=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots $$ and assuming convergence, we have $$ 0=H-2\cdot\frac{1}{2}H=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right) - 2\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)=\\1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\ldots=\ln 2. $$ Since this is a contradiction, the sum must diverge.

Update.

To be more clear about what I'm saying here... I'm not saying that there's no way to assign a value to the harmonic series. Clearly there are any number of ways, some sillier and more arbitrary than others. Call a partial function $S$ from infinite sequences to real or complex numbers a summation method if $S(\{a_i\})$ is defined and equal to $\sum_{i=1}^{\infty}a_i$ whenever the sum converges (in the usual sense); and say that $\{a_i\}$ is $S$-summable if $S(\{a_i\})$ is defined. There are a number of properties that you might want your summation method to preserve from ordinary summation. For instance, you might want it to remain linear: $$S(\{ \alpha a_i + \beta b_i \})=\alpha S(\{a_i\}) + \beta S(\{b_i\})$$ whenever $\{a_i\}$ and $\{b_i\}$ are $S$-summable. You might also want it to remain stable under insertion of zeros: if $J:\mathbb{N}\rightarrow\mathbb{N}$ is increasing and $\{a_i\}$ is $S$-summable, then $S(\{a'_i\})=S(\{a_i\})$, where $a'_{J(i)}=a_i$ and $a'_i=0$ for $i\not\in J^{-1}(\mathbb{N})$. I haven't said anything about analytic continuation or whatever other exotic regularization you want to consider. But what I'm saying, above and in the comments, is that no summation method under which either the harmonic series or $1+2+3+\ldots$ is summable can be both linear and stable.

Now, there's a weaker form of stability (under insertion of only finitely many zeros) that you can preserve while making the harmonic series summable, but not while making $1+2+3+\ldots$ summable. So in that specific sense, the latter series (!) is more pathological. In order to sum the natural numbers, you need to give up either linearity or stability, rendering the manipulations in the original post meaningless.

You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$ (Euler-Mascheroni constant):

$$\lim_{h\to0}\frac{\zeta(1+h)+\zeta(1-h)}2=\gamma$$

The same value can be obtained using the Ramanujan's summation of harmonic series.

So to directly answer your question, yes, we can assign a value to the sum of harmonic series, and at least the two methods give the same result.

Note that the Ramanujan's sum of $1+2+3+4+...$ is $-\frac1{12}$ so this method can be seen to work in both cases.