= Solution
{c}
First we need all equality cases, rather than just the examples from part (c). The threshold estimate in part (b), with $t>0$, shows that any maximum-weight <k-Sperner family> has <Lubell mass> exactly $k$.
For each $A\in\mathcal A$, let $\ell(A)$ be the largest number of members in a strict <chain in a partially ordered set> within $\mathcal A$ ending at $A$. Its values lie in $\{1,\ldots,k\}$. If $A\subsetneq B$ then $\ell(B)\geq\ell(A)+1$, so each height class $\mathcal A_j=\{A:\ell(A)=j\}$ is an <antichain>. Its <Lubell mass> is at most one by the <LYM inequality>, and
$$
k=L(\mathcal A)=\sum_{j=1}^kL(\mathcal A_j).
$$
Consequently every height class has <Lubell mass> one. By <equality in the LYM inequality>, each is a full rank level. Since the classes are disjoint, the maximizing family is a union of exactly $k$ distinct full levels.
Let $t$ again be the $k$th largest $u(i)$, let $h$ ranks have $u(i)>t$, and let $s$ ranks have $u(i)=t$. Equality in the threshold estimate requires all $h$ higher-weight levels, no lower-weight levels, and any $k-h$ of the $s$ tied levels. Thus the <number of maximizing weighted k-Sperner families> is
$$
m=\binom{s}{k-h},\qquad 0\leq h<k,\quad k-h\leq s\leq n+1-h.
$$
Conversely, every such $h,s$ is realizable. Assign $u(i)=2$ on $h$ chosen ranks, $u(i)=1$ on $s$ further ranks, and $u(i)=1/2$ on all remaining ranks, then set $w(i)=u(i)/\binom ni$. These weights are positive and have exactly the stated tie pattern.
Putting $j=k-h$ gives a concise classification of the possible integers:
$$
\boxed{\left\{\binom{s}{j}:1\leq j\leq k,\quad j\leq s\leq n+1-k+j\right\}.}
$$
Repeated values in this set are counted only once. The possible counts need not form an interval: for example, when $n=5,k=2$ they are $1,2,3,4,5,6,10,15$.
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