Negative mass effect? What's wrong with the following argument to "produce" negative mass?:
Consider two charged bodys of weight $m,M$ of distance $r$ with the electric charge $Q$ and $-Q$. Then the energy of this system is:
$$E = (m+M)c^2-k\frac{Q^2}{r}-G \frac{mM}{r},$$ where $k$ is the Coulomb constant. Hence the whole mass of this system is: $$\hat{m} = m+M - \frac{k Q^2+GmM}{rc^2}.$$ If we set this whole mass to $-0.001\text{kg}$ we might solve for $Q$ and get:
$$Q = \sqrt{\frac{-GmM+rc^2(0.001+m+M)}{k} }$$ Now for this to make sense we must have $$r > \frac{G m M }{c^2(0.001+m+M)}.$$ For instance if $r = 0.01 \text{m}$, and $m=M=0.001\text{kg}$, we get $Q = 17.33\text{C} = C \cdot U$. If we let $U=12 \cdot 10^3 \text{V}$, the we get for the capacity of the capacitor: $C= \frac{Q}{U}=0.001444178 \text{F}$. Is this practically achievable? (It would be nice if some experimental physicists would perform an experiment to see if this thought experiment is valid or not)
Edit:
I read about the "positive mass theorem" and it seems like it is based at least on the "null energy condition" which is a conjecture, which is violated by the Casimir effect. In this article, which I do not understand fully, it is investigated what happens if the null energy condition is violated: https://arxiv.org/abs/gr-qc/0003025
It seems like there is no contradiction (if one assumes that the null energy condition can be violated, which is shown by the Casimir effect) to assume there exists negative mass:
Suppose two neutrons are located at distance $r$. Then the whole energy of this system is (by the Yukawa Potential):
$$E = \hat{m}c^2 = 2m_pc^2 - g^2 \frac{e^{-m c r / h}}{r}$$
where $h = 6.58\cdot 10^{-16}$ is the reduced Planck constant.
If we let $$-k m_p = \hat{m} = 2m_p - g^2 \frac{e^{-m c r / h}}{rc^2}$$ where $k=10^{27}$ and $m_p$ is the mass of the neutron, hence $\hat{m}=-1.67 \text{kg}$ then solving for $r$ (for example with wolfram alpha), we get:
$$r = \frac{h W(\frac{g^2m}{2chm_p+chkm_p})}{cm},$$ where $W$ is the Lambert W-function. If we set $g=m=1$, which I did, because I am not comfortable with the values of $g$ and $m$, we get:
$$r = \frac{h W(\frac{1}{2chm_p+chkm_p})}{c} \equiv 4.14 \cdot 10^{-20} \text{m},$$ which is still greater then the Planck length.
So, is it possible to bring two neutrons this near?
Second edit:
The thought experiment with the differently charged bodys might be thought of as charging a capacitor. It is shown here experimentally, that it is possible to reduce the mass of the capacitor when charging it:
https://arxiv.org/abs/1502.06915. 
This effect is called the Biefeld-Brown effect. It would be interesting to calculate if this thought experiment can explain the Biefeld-Brown effect.
Update for those interested:
It seems that this concept of the first thought experiment has already been explored by others:
https://tu-dresden.de/ing/maschinenwesen/ilr/rfs/ressourcen/dateien/forschung/folder-2007-08-21-5231434330/ag_raumfahrtantriebe/JPC---Propellantless-Propulsion-with-Negative-Matter-Generated-by-Electric-Charges.pdf?lang=en
The first thought experiment can be dated back to:
http://aapt.scitation.org/doi/citedby/10.1119/1.11328
 A: If $E = (m+M)c^2+k\frac{Q^2}{r}-G \frac{mM}{r} < 0$, then $(m+M)c^2-G \frac{mM}{r}$ must also be less than zero. Thus we must have $r < \frac{GmM}{(m+M)c^2}$.
This is less than the Schwarzschild radius of a particle of mass $m+M$, and this gets to the heart of the matter: You are using Newtonian gravitation in a field where it does not apply.
A: You already have several good answers. In short, the error lies in (inconsistently) mixing Newtonian gravitation and relativistic effects.
That being said, one may wonder what general relativity has to say. In other words, we already know what the naïve picture predicts; now we want to know what the more correct one does.
In full generality, the answer is given by the Positive energy theorem. This theorem proves that, under some natural assumptions, the mass of any system is always positive (or null, for flat spacetime). In this sense, one would always have $M>0$ for physical situations. General relativity predicts that mass cannot be negative. Neat, right?
Let me illustrate this theorem through a couple of (relevant) examples. Your description corresponds to a two-body problem, for which no exact solution is known. It can nevertheless be approximated by a one body problem by taking one of the masses to be much larger than the other one, which we assume from now on. Under this circumstance, you can describe your charged system as a Reissner-Nordström black hole. This black hole is essentially the same as the Schwarzschild one, but with a non-zero electric charge. Its mass is given by
$$
M=M_0+\frac{Q^2}{16\pi GM_0}
$$
where $M_0$ is the mass it would have for $Q=0$. Here we see that $M>0$ for any value of $Q$, in accordance with the theorem above.
A more general situation is given by the Kerr-Newman one, where we also have a non-zero angular momentum (the BH is rotating). Here the expression for $M$ is more complicated, but it can be shown that it is also positive (in fact, it is larger than the Reissner-Nordström one, which is itself larger than $M_0$; this is easy to understand: charge adds energy, and rotation adds even more energy).
Interestingly, it can also be shown that in both cases $Q$ cannot be too large, or otherwise the solution becomes pathological. In other words, even if $M$ is always positive, if you keep increasing $Q$ (or the angular momentum in the KN case), at some point the black-hole starts violating basic principles (naked singularities, which are no good). To some extent, this is the analogous situation to your "negative mass" result, but taking into account relativistic effects. For a very large value of $Q$ the model does also break down, but for different reasons.
A: Following your reasoning, you don't need any electric charge to obtain a negative energy $E$ and with $E=mc^2$ also a "negative mass" $\hat{m}$ if the distance $r$ between your masses becomes small enough: $$E = \hat{m} c^2=(m+M)c^2-G \frac{mM}{r}$$ 
