= Solution
Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the <Second law of thermodynamics>, equilibrium maximizes the total <entropy>
$$
S_{\rm tot}(E_1)=S_1(E_1)+S_2(E-E_1).
$$
At an interior maximum,
$$
0=\frac{dS_{\rm tot}}{dE_1}
=\frac{\partial S_1}{\partial E_1}
-\frac{\partial S_2}{\partial E_2}
=\frac1{T_1}-\frac1{T_2}.
$$
Thus $T_1=T_2$, which is <entropy maximization under thermal contact> and the condition for <thermal equilibrium>.
Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Since
$$
\frac{\partial^2S}{\partial E^2}
=\frac\partial{\partial E}\left(\frac1T\right)
=-\frac1{T^2}\frac{\partial T}{\partial E}<0,
$$
one has $\partial T/\partial E>0$, or equivalently
$$
\boxed{\frac{\partial E}{\partial T}>0.}
$$
This is the <entropy concavity and positive heat capacity> criterion.
Now let $N_\uparrow=\alpha N$ and $N_\downarrow=(1-\alpha)N$. Because the spins are distinguishable, the number of microstates is the <binomial coefficient>
$$
\Omega=\binom{N}{\alpha N}
=\frac{N!}{(\alpha N)!((1-\alpha)N)!}.
$$
The <Boltzmann entropy> $S=k_B\log\Omega$ and the <Stirling formula> give, to leading order for large $N$,
$$
\boxed{
S(\alpha)=-Nk_B\left[\alpha\log\alpha
+(1-\alpha)\log(1-\alpha)\right].
}
$$
The total energy is
$$
\boxed{E(\alpha)=N\bigl(\alpha\epsilon-(1-\alpha)\epsilon\bigr)
=N\epsilon(2\alpha-1).}
$$
Therefore
$$
\frac1T=\frac{dS/d\alpha}{dE/d\alpha}
=\frac{k_B}{2\epsilon}\log\left(\frac{1-\alpha}{\alpha}\right).
$$
Solving for the up-spin fraction gives the <independent spin-one-half two-level system> result
$$
\boxed{
\alpha(T)=\frac1{1+e^{2\epsilon/(k_BT)}}.
}
$$
For ordinary positive absolute temperature,
$$
\boxed{0\leq\alpha<\frac12,}
$$
with $\alpha\to0$ as $T\to0^+$ and $\alpha\to1/2$ as $T\to+\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha\leq1$ is also mathematically possible and corresponds to <negative temperature>; $\alpha=1/2$ is the infinite-temperature state.
Substitution gives
$$
E(T)=-N\epsilon\tanh\left(\frac\epsilon{k_BT}\right),
$$
and direct differentiation yields
$$
\boxed{
\frac{dE}{dT}
=\frac{N\epsilon^2}{k_BT^2}
\operatorname{sech}^2\left(\frac\epsilon{k_BT}\right)>0.
}
$$
\b[Thus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.]
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