= Solution
Write $x=\Delta E/(k_{\rm B}T)$ and $r=N_\uparrow/N_\downarrow=e^{-x}$. Since $N=N_\uparrow+N_\downarrow$, eliminate the individual populations:
$$
\frac{N_\uparrow-N_\downarrow}{N}=\frac{r-1}{r+1}=\frac{e^{-x}-1}{e^{-x}+1}=-\tanh\frac x2.
$$
The <Taylor expansion> $e^{-x}=1-x+O(x^2)$ gives the same result, and the odd expansion of the <hyperbolic tangent> yields
$$
\boxed{\frac{N_\uparrow-N_\downarrow}{N}=-\frac x2+O(x^3)\simeq-\frac{\hbar\gamma B}{2k_{\rm B}T}.}
$$
Equivalently, $N_\uparrow\simeq(N/2)(1-x/2)$ and $N_\downarrow\simeq(N/2)(1+x/2)$. Their sum is still $N$. The negative sign follows directly from the stated higher-energy up-state population, and does not mean a negative magnitude of the equilibrium magnetization.
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