= Solution
Define neutral-hydrogen <column density> along a ray by $N=\int n_{\rm HI}\,d\ell$, in atoms per unit area. The <neutral-hydrogen column-density distribution> per redshift is
$$
f_z(N,z)=\frac{\partial^2\mathcal N}{\partial N\,\partial z},
$$
so $f_z\,dN\,dz$ is the expected number of absorbers in those intervals along one sightline. A logarithmic column bin instead has distribution $N\ln(10)f_z$; the bin convention must be specified.
Let $n_c$ be a constant <comoving number density> and $\sigma_p$ a constant physical interception area. The corresponding physical <number density> is $n_p=n_c(1+z)^3$. Proper light-path length satisfies $d\ell=c|dt|=c\,dz/[(1+z)H(z)]$. The expected intersection count, or first-order probability in infinitesimal $dz$, is therefore
$$
\frac{d\mathcal N}{dz}=n_p\sigma_p\frac{d\ell}{dz}
=\frac{c n_c\sigma_p(1+z)^2}{H(z)}.
$$
The <Friedmann equation>, the <constant-equation-of-state density scaling> for <pressureless matter>, and the present <cosmological density parameters> give
$$
H^2(z)=H_0^2\left[\Omega_m(1+z)^3+\Omega_\Lambda+\Omega_K(1+z)^2\right],
\qquad\Omega_m+\Omega_\Lambda+\Omega_K=1.
$$
Combining them yields the <absorber incidence and comoving number density> relation
$$
\boxed{\frac{d\mathcal N}{dz}\propto
\frac{(1+z)^2}{\sqrt{\Omega_m(1+z)^3+\Omega_\Lambda+\Omega_K(1+z)^2}}.}
$$
Here the constant cross-section is physical, as required for this result; a fixed comoving cross-section would instead decrease in physical area as $(1+z)^{-2}$. It is convenient to define <absorption distance> by $dX/dz=(H_0/H(z))(1+z)^2$ and $f_X=f_z/(dX/dz)$, so a population with these non-evolution assumptions has constant incidence per $X$.
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