Magnetic properties of electrically charged metal Assume that I have two identical pieces of metal (let it be iron). I negatively charge one of them. Then I magnetize both of them. Which piece would have stronger magnetic field? I am guessing the electrically charged one should have stronger magnetic field, since magnetizm is a property of electrons and negatively charged piece would have more electrons. 
Similarly, if we take two identical pieces of metal and positively charge one of them, then magnetize both. The positively charged piece will have weaker magnetic field, since there are few electrons? 
Is my guess accurate? 
 A: I don't think your description is accurate. The magnetic property of a material is not determined by the number of electrons in total, but rather by (a) the number unpaired magnetic moments, and (b)  by the interaction between magnetic domains in the solid.
First, let's consider a single atom. It is composed of a nucleus and multiple electrons. Each electrons possess a magnetic moment $\vec \mu$ which is associated with the spin via $\vec \mu = - g_s \mu_{Bohr} \vec{S}/\hbar$ where $g_s \approx 2$. Furthermore, each electron occupies a "different" energy level of the atom. These energy levels (or electric orbits) are usually denoted by $\{1s, 2s, 2p, 3s, 3p, 4s, 3d, \ldots \}$ and filled up starting from the energetically lowest state -- note that each state can be occupied by one "spin up" and one "spin down" electron. 
Due to interactions between the electrons different spin configurations differ in energy. How these energy levels are filled up in detail is described by Hund rule. However, one thing to notice is that half filled and fully filled configurations possess "low" energies. Hence, they are favourable. 
So let's look at the electronic configuration of an iron atom, 
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$. As you probably know, the completely filled states are uninteresting. Thus, we concentrate on the valence electrons, $3d^6$. This means iron has six electrons in the $d$-orbit, which could im principle be occupied by 10 electrons, 
$$
L = 2
\Rightarrow m_l = 0, \pm 1, \pm 2 
$$
and
$$
S = \frac{1}{2}
\Rightarrow  m_s = \pm \frac{1}{2}
$$ 
Thus, an iron atom has four unpaired electrons

Note that adding electrons to an iron atom, would not result in an increase in the number the unpaired electrons, but a decrease. 
Now, your question is not about the magnetic properties of $Fe^+$ and $Fe^-$ atoms, but about a bulk material. Therefore, we can not stop at the atomic level, but need to account for the solid. I'm not an expert in solid state physics, but I know that magnets are build from magnetic domains. These are not all parallel -- things are messy in solid state physics. 
Skipping all this messiness, I'm pretty sure that additional electrons will not alter the magnetic property of iron significantly. Neither does the subtraction of electrons. My argument is that iron is an electric conductor. Thus, the crystalline structure does not localise  all electrons around the nuclei, but forms a so called "electron gas". The electrons of the "electron gas" are not locked in the band structure of the solid, but are free to move around. While they are moving, they interact we the solid (nuclei, bounded electrons, and unbounded electrons). Since the magnetic interaction is "much weaker" than the electric interaction, I do not expect that the electrons do maintain their spin orientation. This implies that they not significantly change the total magnetic moment.
