= Solution
Substitute harmonic amplitudes with the specified $e^{-i\omega t}$ convention:
$$
m(\omega_0^2-\omega^2-i\gamma\omega)\mathbf u_0=q\mathbf E_0.
$$
One electron contributes a dipole moment $\mathbf p=q\mathbf u_0$, so the <electric polarization> of a density $n$ is
$$
\mathbf P_0=nq\mathbf u_0
=\frac{nq^2}{m(\omega_0^2-\omega^2-i\gamma\omega)}\mathbf E_0.
$$
Using $\mathbf D=\epsilon_0\mathbf E+\mathbf P=\epsilon_0\epsilon_r\mathbf E$ gives
$$
\boxed{\epsilon_r(\omega)=1+\frac{\omega_p^2}{\omega_0^2-\omega^2-i\gamma\omega},\qquad
\omega_p^2=\frac{nq^2}{m\epsilon_0}.}
$$
Thus the paper's dimensionless $\epsilon_\omega$ is the <relative permittivity>, rather than the absolute <permittivity> $\epsilon_0\epsilon_r$. The electron's negative charge disappears from the <electric susceptibility> because both the force and dipole moment contain $q$.
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