How to justify "correct Boltzmann counting" (dividing by $N!$) without resorting to quantum mechanics? Let a box of volume $V$ contain $N$ monatomic gas particles, with total energy $E$. By considering the "volume" of phase space consistent with our macro-observables ($E$, $V$, and $N$), the multiplicity of this macrostate is (from Ref.1)
$$
\Omega = \frac{V^N}{h^{3N}}\frac{2\pi^{3N/2}(2mU)^{(3N-1)/2}}{\Gamma(3N/2)} \,.
$$
However, this expression is "incomplete" as it leads to a non-extensive entropy and hence Gibbs paradox. The fix for this is dividing by $N!$ (or "correct Boltzmann counting"), and the justification is usually: because particles are fundamentally indistinguishable in quantum mechanics.
I don't like the quantum mechanical justification, and Ref.2 sums up my feelings

If this reasoning were correct, then the Gibbs paradox would imply
  that the world is quantum mechanical.  It is difficult to believe that
  a thought experiment from classical physics could produce such a
  profound insight.

Furthermore, I intuitively accept that the probability of a macrostate is proportional to the volume of phase space it covers. Dividing by $N!$ would "unfairly" reduce the probability of macrostates with larger number of particles. If $E$, $V$, and $N$ were all doubled, then the compliant region of phase space would increase drastically, more than just getting squared (i.e. doubling in entropy).
On the other hand, if entropy is not an extensive quantity, then neither is chemical potential $\mu$, and using the earlier "uncorrected" multiplicity, an infinite particle reservoir at constant temperature and pressure will have non-finite chemical potential, this can't be correct. (Related to: How do you experimentally measure the chemical potential of a gas?)
According to Ref.2,

systematic division by $N!$ for all particles of the same kind in
  mutually accessible states has no empirical consequences.

and it goes on to say that this is because dividing by $N!$ in closed systems do not affect the relative probabilities of macrostates (as $N$ is constant).
My understanding didn't really "click" after reading Ref.2, so I suppose I'm looking for a more detailed (or "intuitive") argument as to why correct Boltzmann counting has no empirical consequences.



*

*Schroeder, D. (1999). An Introduction to Thermal Physics. p.71.

*Versteegh, M., Dieks, D. (2011). "The Gibbs Paradox and the distinguishability of identical particles".

This question is similar to Why is the partition function divided by $(h^{3N} N!)$?, but I'm looking for a non quantum mechanical explanation specifically.
 A: There is no need for quantum mechanics. You can find a classical argument in Landau-Lifshitz, Statistical Physics, Chapter III, Paragraph 31.
What you have to understand is that when integrating over phase space, for example to calculate the free energy in the canonical ensemble, you should not integrate over all values of the canonical variables $p,q$, but only over the values which correspond to different physical states. Landau uses the notation $\int'$ to denote this; for example for the free energy you will have
$$\beta F = -  \ln \int' e^{-\beta \mathcal H(p,q)} dp dq$$ 
where $\beta=1/k_B T$ as usual. 
Now, it is often more practical to just integrate over the whole phase, i.e. to take $\int dp dq$ space instead that over the portions of phase space corresponding to different states, i.e. $\int' dp dq$. 
However, if you do so, you are overcounting the number of states. In the case of a gas of $N$ atoms, in order to get the correct counting, you have to divide the integral extended over the whole phase space by the number of possible permutations of $N$ atoms, which is $N!$. This is because if you switch the position of two identical atoms you are not getting a new physical state (this is the key point of the whole argument). To sum up, for a gas of $N$ identical atoms we have
$$\int ' \dots dp dq = \frac 1 {N!} \int \dots dp dq$$
which is Eq. 31.7 in my edition of Landau-Lifshits.
A: This is a great question and both you and @valerio have given good answers.  I will just add that, as suggested in the Jaynes paper you cite, the Gibbs paradox is really confusion about macrovariables and what constitutes a "subsystem".  Entropy is only ever defined with respect to some set of macrovariables, so if you want to compare the total entropy with the entropy of the subsystems, you need to use the same macrovariables for the subsystems as for the complete system.
Specifically, if the particles are indistinguishable, $V_i$ is not a macrovariable.  This is obviously true with the partition removed, since there is no way to tell from macro observations (that is, external interactions) which particles are in volume $i$.  Less obviously, it is also true with the partition in place, since "indistinguishable" means there's still no way to tell from macro observations which particles are in volume $i$.  The only macrovariable related to volume is the total volume, $V$.
Meanwhile, a meaningful way to divide the system into subsystems is to split it into $k$ groups of particles with $\{N_1,...,N_k\}$ particles in each group, where $N = \sum N_i$.  (It's helpful to realize these particles need not be physically grouped - say, on one side of the box or the other.  You could, for example, make a subsystem by counting off every third particle.  Importantly, the subsystem is not identified with a volume of space, but with a set of particles.)  Then by "extensive entropy", we mean that $S(N,V) = \sum S_i(N_i,V)$.  That is, in this scheme you should think of the $i$th subsystem as occupying the entire volume $V$, as you indicated in your "cool observation".  It doesn't matter if it happens that all the particles in a subsystem are grouped into a smaller volume like the left side or the right side or the upper right corner; such a fortuitous configuration is a particular microstate of the subsystem, for which entropy is not defined.
(As for $E$, we must choose the subsystems such that they are all in mutual thermodynamic equilibrium, so $E_i/N_i \approx E/N \,\forall\, i$.  To say it another way, the correct macrovariable is not the energy but the temperature.)
Now the statement that the entropy is extensive can be written
$ S(T,N,V) = \sum S_i(T,N_i,V) $, or $\Omega(T,N,V) = \prod \Omega_i(T,N_i,V)$.
That is trivially true for an ideal gas of identical particles, however you count the states.
How does this change if the particles are distinguishable?  Well, "distinguishable" means there is some macrovariable whose value depends on the states of the different types.  If there is a mixture of xenon and neon atoms in the box, there is the possibility of introducing new macrovariables IF the setup includes some mechanism that reacts to xenon and neon differently, e.g. a partition that is permeable to neon but not to xenon.  By inserting the partition, the relative fraction of xenon on the left and on the right are now available as macrovariables, since a difference in xenon concentration across the partition will produce a force on the partition.  Furthermore, you could use the force to do work on the exterior or exert work on the system through the displacement of the partition, indicating that state differences involving changes in the new macrovariables are indeed associated with changes of entropy.  You might quantify the new macrovariables as $N_{xl}/V_l$ and $N_{xr}/V_r$, or equivalently, $N_{xl}$, $V_l$, and $N_x$ (in addition to $V$, which is still a macrovariable.)  So in terms of these macrovariables, $S = S(T,N,V,N_x,N_{xl},V_{xl})$ for the entire system AND for the subsystems.
