= Solution
Continue with $0<h<d$ and set $r=d/h$. Part (e) requires an interior count to satisfy $s\leq r\leq s+1$ with $s\geq k$; the all-deer endpoint $s=n$ is allowed when $r\geq n$. If $r$ is nonintegral and $k\leq r<n$, only $s=\lfloor r\rfloor$ works. If $r=m$ is an integer with $k+1\leq m\leq n$, both $s=m-1$ and $s=m$ work. At $r=k$ only $s=k$ remains, while $r>n$ permits only $s=n$. When $r<k$, there is no positive equilibrium count.
Thus there are \b[at most two positive counts], and \b[exactly one] precisely when
$$
\boxed{\frac dh\geq k\quad\text{and}\quad\frac dh\notin\{k+1,k+2,\ldots,n\}.}
$$
This includes $k=n$, when the excluded set is empty. It counts numbers of deer hunters, not the many choices of which hunters participate. The argument is the <threshold stag-hunt equilibrium count>.
The original paper states only $h<d$ explicitly. Its assertion needs the usual strict positivity of the hare reward. If zero is included, take $n=6,k=4,d=1,h=0$. Counts $s=1$ and $s=2$ both give unsuccessful hunters zero, and one additional deer hunter still cannot reach the threshold; all unilateral alternatives also pay zero. The count $s=6$ is an equilibrium as well. These are \b[three positive equilibrium counts], contradicting an unrestricted reading. More generally, for $h=0<d$, the positive counts are $1,\ldots,k-2$ and $n$.
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