Why is there a finite current density but zero free charge density inside a resistor in DC? If we take a (lets imagine cylindrical) resistor in DC (steady state), we have that the electric field follows Ohm’s law:
$\mathbf J_f=\sigma  \mathbf E$. where $\mathbf J_f$ is the free electron current density.
Since it is in steady state it also follows from the continuity equation that $\nabla \cdot \mathbf J_f=0$
Putting the first equation inside the second we get for an homogeneous medium that: $\nabla \cdot (\sigma  \mathbf E)=0$, hence $\nabla \cdot \mathbf E=0$, hence according to Gauss’ law $\nabla \cdot \mathbf E= \rho_f/\epsilon =0$.
In other words in steady state the free charge inside a resistor is zero.
The big problem is that according to any book I have read (although not a mathematical reason have been given) charge density and electric field are spatially uniform inside a resistor in DC.
Yet,  $\mathbf J=\rho_f  \mathbf V$ (where $\rho_f $ is the free charge density), and since $\rho_f=0$ , $\mathbf J$ and $\mathbf E$ should be zero
How can $\mathbf E$ and $\mathbf J$ be nonzero and uniform in steady state, if the above equations indicate they should be zero?
 A: 
Yet,  $\mathbf J=\rho_f  \mathbf V$ (where $\rho_f $ is the free charge density), and since $\rho_f=0$ , $\mathbf J$ and $\mathbf E$ should be zero

The problem that you are running into is that this statement is not really correct. It should actually be: $\mathbf J_f= \Sigma \rho_i \mathbf V_i$ (where $\rho_i$ is the charge density of the i’th type of free charge and $\mathbf V_i$ is its drift velocity).
In a typical metallic conductor you will have a very large negative $\rho_{electron}$ with a very small $\mathbf V_{electron}$. You will have an equally large positive $\rho_{proton}$ with $\mathbf V_{proton}=0$. Note that even though the protons are fixed in place they are still considered free charges. This is because they do not form dipoles with an overall neutral charge. Bound charges are overall neutral but with a dipole moment that can be polarized.
In an electrolyte like a sodium chloride solution you will have a negative $\rho_{Cl^-}$ and an equal magnitude but positive $\rho_{Na^+}$ each with their own velocities pointed in opposite directions.
Since different types of free charge will have different velocities you cannot simply lump them all together as your expression tried to do.
A: $\rho$ is zero inside a resistor, because the positive and negative charges cancel each other. $J$ is non-zero since only electrons do the moving.
A: You have mis-stated Gauss's Law.
Gauss's Law can be written as either
$${\bf\nabla}\cdot{\bf E}=\frac{\rho}{\epsilon_0}$$
where $\rho$ is the total charge (not the free charge); or as
$${\bf\nabla}\cdot{\bf D}=\rho_f$$
where $\bf D$ is the electric displacement field and $\rho_f$ is the free charge.
So we don't have 0 free charge (if we did, $\sigma$ would be zero), we have 0 total charge, including both free (current carriers) and fixed charges (nuclear protons).
A: In a current carrying conductor (with any amount of resistance), the free electron density and the total charge density is not constant. A power supply takes electrons from the positive end and puts them into the negative end. In a uniform conductor a uniform current requires a uniform electric field which implies a uniform gradient in the charge density.
Gauss's law tells us that a uniform field can be produce by one or more infinite sheets of charge. Clearly, we don't have that in electric circuit. So the question becomes; what kind of charge distribution can maintain a nearly uniform field within a length of wire which can have multiple bends, curves or loops?  Each short segment of wire must see more charge beyond one end than beyond the other.  This requires a non-uniform distribution of charge along the wire.  Then each such segment (except one near the center) must contain a net charge.  But the flux coming in from one end must equal that leaving at the other. This means that a segment with a net + charge must have flux leaving through the side of the wire, and one with a negative charge will have flux entering through the side. These components will participate in the divergence of the field, and will cause a radial shift in the distribution of free electrons.
