= Solution
For a one-dimensional free-electron gas of length $L$, allowed <wave vectors> have spacing $2\pi/L$. Including two <spin> states and occupied <wave vectors> from $-k_F$ to $+k_F$,
$$
N_e=2\frac{L}{2\pi}(2k_F),\qquad
\boxed{k_F=\frac{\pi n_e}{2},\qquad
E_F=\frac{\hbar^2k_F^2}{2m_e}=\frac{\pi^2\hbar^2n_e^2}{8m_e}.}
$$
The ground-state <kinetic energy> is found by integrating rather than assigning every <electron> the <Fermi energy>:
$$
K_e=2\frac{L}{2\pi}\int_{-k_F}^{k_F}\frac{\hbar^2k^2}{2m_e}\,dk
=\frac{L\hbar^2k_F^3}{3\pi m_e}
=\boxed{\frac13N_eE_F.}
$$
<Charge neutrality> gives $N_e=ZN$ because each positive ion has charge $+Ze$. Thus $K_e=ZNE_F/3$. Since the ion separation is $2R$, $L\simeq2RN$ and $n_e=Z/(2R)$. These are one-dimensional number densities and state counts; using a three-dimensional $k_F\propto n^{1/3}$ relation here would change the model. The results concern the ground-state Fermi-gas contribution, neglecting <electron> interactions beyond the supplied electrostatic <energy>.
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