= Solution
Assume $eB>0$; reversing the sign only exchanges the spin labels. Since $\mathbf B\mathbin{\cdot}\boldsymbol\sigma=B\sigma_3$, the Pauli term shifts the spinless energies by $\pm\hbar\omega_c/2$. Thus
$$
E_{n,\sigma}
=\hbar\omega_c\left(n+\frac12+\frac\sigma2\right),
\qquad \sigma=\pm1.
$$
After collecting equal energies, the <spin splitting of Landau levels> is
$$
\boxed{E_\ell=\ell\hbar\omega_c},
\qquad \ell=0,1,2,\ldots.
$$
The level $\ell=0$ contains only $(n,sigma)=(0,-1)$ and has degeneracy $D$. Every level $\ell\geq1$ contains $(\ell,-1)$ and $(\ell-1,+1)$ and has degeneracy $2D$.
For $N$ noninteracting electrons, the <Pauli exclusion principle> fills these states from the bottom. If $0\leq N\leq D$, every electron fits in the zero-energy level and
$$
E_{\mathrm{gs}}(N)=0.
$$
For $N>D$, write
$$
N-D=2Dk+r,
\qquad k\in\mathbb Z_{\geq0},
\qquad 0\leq r<2D.
$$
Then the first $k$ positive levels are full and $r$ states in level $k+1$ are occupied. The <ground state>[ground-state] energy is
$$
\boxed{
E_{\mathrm{gs}}(N)
=\hbar\omega_c\left[Dk(k+1)+(k+1)r\right].
}
$$
Equivalently, on
$$
D(2k+1)\leq N\leq D(2k+3),
$$
$$
E_{\mathrm{gs}}(N)
=D\hbar\omega_c k(k+1)
+(k+1)\hbar\omega_c\,[N-D(2k+1)].
$$
The graph is continuous and piecewise linear: it is flat up to $N=D$, then has slopes $\hbar\omega_c,2\hbar\omega_c,3\hbar\omega_c,\ldots$, with kinks at
$$
\boxed{N=D,3D,5D,\ldots.}
$$
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