= Solution
Take two localized <spin> orientations with moments $\pm\mu_B$ along the applied <magnetic field>. This assumes an electron-like spin $g$ factor of two. Their <energies> are $\mp\mu_BB$, so the one-moment <partition function> is $Z_1=2\cosh x$, with $x=\mu_BB/(k_BT)$. The thermal mean moment is $k_BT\,\partial_B\ln Z_1=\mu_B\tanh x$, and the <spin-one-half paramagnetic magnetization> is
$$
\boxed{M=n\mu_B\tanh\!\left(\frac{\mu_BB}{k_BT}\right).}
$$
For $|x|\ll1$, $M\simeq n\mu_B^2B/(k_BT)$. With the same SI convention $\chi=\partial M/\partial H$ and $B\simeq\mu_0H$,
$$
\boxed{\chi=\frac CT,\qquad C=\frac{n\mu_0\mu_B^2}{k_B}.}
$$
This <Curie law> is a zero-field or weak-field susceptibility. At arbitrary field the differential response is $(C/T)\operatorname{sech}^2x$, and $M$ saturates at $n\mu_B$; extending the linear law to unlimited field would contradict that finite <saturation magnetization>.
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