= Solution
For <state counting in a Bloch band>, take a <primitive cubic lattice> of spacing $a$ and impose <periodic boundary conditions> on lengths $L_i=N_i a$. The allowed <wavevectors> are $k_i=2\pi l_i/L_i$. The first <Brillouin zone> has side $2\pi/a$, so it contains $N_i$ distinct values in each direction and hence $N_xN_yN_z=N$ wavevectors. Each orbital-derived <energy band> has one state per wavevector per <spin> orientation. Therefore
$$
\boxed{\text{states per band in the whole crystal}=2N,\qquad
\text{states per atom per band}=2.}
$$
The phrase “per atom” attached to $2N$ in the PDF is a counting error: $2N$ is the total, not the per-atom count.
By the <Pauli exclusion principle>, a monovalent crystal supplies $N$ valence <Electrons> and half-fills a spin-degenerate band; its <Fermi level> lies within allowed states, giving a <metal> in the independent-electron picture. Two electrons per atom can fill one band. If a <band gap> separates it from all empty bands, the crystal is a <band insulator>. An even valence count only permits insulation: overlapping bands can still produce a <metal>. Conversely, electron interactions can produce a <Mott insulator>, and symmetry breaking can permit insulation outside this simple spin-degenerate <electronic band structure> criterion. The relevant count in a crystal with a basis is per <primitive unit cell>.
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