a) In the Milne frame the spatial metric of the slice $\tau=const$ is
\[dl^{2}_{\tau}
=\tau^2 dr^2 +\tau^2 \sinh^{2}r d\Omega^2,\]
so the volume element is (integrating over $d\Omega$)
\[dV_{\tau}=4\pi \tau^{3}\sinh^{2}r\, dr.\]
If $n_{\tau}=const$ is the particles' number density, the total number of particles within a sphere of radius (radial coordinate, to be precise) $r_{\star}$ is
\[N_{\tau}(r_{\star})=4\pi\;
\underbrace{n_{\tau}(\tau)\tau^{3}}_{const}
\int\limits_{0}^{r_{\star}}dr\,\sinh^{2}r
\xrightarrow[r_{\star}\to\infty]{}\infty,\]
which is natural, as we assume the number density is constant at a given time on an unbounded hyperboloid of infinite volume.
b) In the Minkowski frame the spatial metric of the slice $T=const$ is
\[dl^{2}_{T}=dR^{2}+R^{2}d\Omega,\]
so the volume element is $dV_{T}=4\pi R^2 dR$. We can rewrite it in terms of $r$, while keeping in mind that we are talking of the slice $T=const$, and thus from (\ref{Milne-Mink}) $R=T\tanh r$ and $dR=Tdr/\cosh^{2}r$ we get
\[dV_{T}=4\pi R^{2}dR
=4\pi T^{3}\frac{\sinh^{2}r\,dr}{\cosh^{4}r}
=4\pi \tau^{3}\frac{\sinh^{2}r\,dr}{\cosh r}.\]
As $r$ is the comoving coordinate, and the number of particles is conserved, it stays the same in the layer between $r$ and $r+dr$ regardless of the chosen spatial slice, so
\[dN[r,dr]=n_{\tau}dV_{\tau}=n_{T}dV_{T}\]
and
\[n_{T}=n_{\tau}\cosh r =n_{\tau}\gamma
=\frac{n_{\tau}}{\sqrt{1-R^2 /T^2}}.\]
We could have obtained this result by simple observation that the number (or mass) density in the Minkowski frame changes due to the contraction of the radial distance between the particles by the Lorentz-factor $\gamma$.
However, when considering the slice $T=const$, we now also have to take into account that $\tau$ is a function of $R$:
\[\tau^{2}=T^{2}-R^{2},\]
so due to $n_{\tau}\tau^{3}=const$, we get
\begin{equation}\label{Milne-density}
n_{T}=n_{0}\frac{(T_{0}/T)^{3}}
{\big[1-R^2 /T^2\big]^{2}},
\end{equation}
which leads to the correct divergence of the integral for the total number of particles on the horizon $R=T$:
\begin{align*}
N_{T}&=\int n_{T}dV_{T}
=4\pi\;n_{0}T_{0}^{3} \int\limits_{0}^{T}
\frac{R^2\,dR /T^3}{(1-R^2 /T^2)^{2}}=\\
&= \Big\| \xi=R/T\;\;\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{1}
\frac{\xi^{2}\,d\xi}{(1-\xi^2)^{2}}=\\
&=\Big\| \xi=\tanh r\Big\|
=4\pi\;n_{\tau}\tau^{3} \int\limits_{0}^{\infty}
\sinh^{2}r\,dr\to\infty.
\end{align*}
In terms of mass density the divergence of $\rho_{T}$ at the horizon means that even for arbitrarily small density $\rho_{0}$ in the comoving frame the density in the laboratory frame becomes arbitrarily large close enough to the horizon. Thus in the neighborhood of the horizon we cannot actually consistently neglect the mass distribution and the curvature of spacetime it induces, and even for small $\rho_0$ we cannot define the world time $T$ in all of the originally Minkowski spacetime.