= Solution
Because $S_{\rm tot}^z=N/2-N_f$, the exact finite-size answer is $m_N=1/2-N_f^{\rm gs}/N$, with $N_f^{\rm gs}$ determined by the parity-consistent filling above. For the thermodynamic <magnetization of the quantum XX chain>, the occupied fraction is the <length> of the negative-energy interval divided by the Brillouin-zone <length>:
$$
\frac{N_f}{N}\longrightarrow\frac1{2\pi}\int_{-k_F}^{k_F}dk
=\frac{k_F}{\pi}.
$$
Therefore
$$
\boxed{m_z(h)=\begin{cases}
\displaystyle\frac12-\frac1\pi\arccos(h/J)
=\frac1\pi\arcsin(h/J),&0\le h\le J,\\[3pt]
\displaystyle\frac12,&h\ge J,
\end{cases}\qquad h_c=J.}
$$
The zero-field limit has zero <magnetization> and the high-field limit has all <spins> up. The finite-ring curve consists of <magnetization> steps approaching this continuous limit. Strictly above $h_c$ the fully polarized <ground state> is unique; exactly at $h_c$ a zero-energy spin-flip state can be degenerate with it. This finite-size qualification does not alter the saturation field or limiting <magnetization> density.
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