= Solution
For $1\le r\le n$ and a rank-$r$ <uniform set family> $\mathcal F$, let $\partial\mathcal F$ be its <lower shadow>. Count pairs $(B,A)$ with $A\in\mathcal F$, $B\subset A$ and $|B|=r-1$. Every $A$ contributes $r$ pairs, whereas each shadow member lies in at most $n-r+1$ rank-$r$ sets. Thus
$$
r|\mathcal F|\le(n-r+1)|\partial\mathcal F|,
\qquad
\boxed{\frac{|\partial\mathcal F|}{\binom n{r-1}}\ge
\frac{|\mathcal F|}{\binom nr}.}
$$
This is the <Local LYM inequality>. Complementation gives the upper-shadow form $|\nabla\mathcal F|/\binom n{r+1}\ge|\mathcal F|/\binom nr$ for $r<n$.
To deduce the <LYM inequality>, let $\mathcal A$ be an <antichain>, $\mathcal A_r$ its rank-$r$ part, and $U_r$ all rank-$r$ sets containing some member of $\mathcal A$. For $r\ge1$,
$$
U_r=\nabla U_{r-1}\ \dot\cup\ \mathcal A_r.
$$
The union is disjoint because an <antichain> member cannot contain a smaller member; every other set of $U_r$ contains a rank-$(r-1)$ superset of its smaller witnessing member. Apply the upper local inequality and put $u_r=|U_r|/\binom nr$:
$$
u_r\ge u_{r-1}+\frac{|\mathcal A_r|}{\binom nr}.
$$
Since $u_0=|\mathcal A_0|$ and $u_n\le1$, summing gives
$$
\boxed{\sum_{r=0}^n\frac{|\mathcal A_r|}{\binom nr}\le1.}
$$
Thus the deduction uses the local inequality explicitly, rather than quoting a separate maximal-chain argument.
Let $M=\binom n{\lfloor n/2\rfloor}$, the largest rank size. The LYM sum is at least $|\mathcal A|/M$, proving the <Sperner theorem> bound. If $|\mathcal A|=M$, equality forces all its members onto ranks with <binomial coefficient> $M$. For even $n$, there is only the middle rank, so the family is that whole rank.
For $n=2m+1$, only ranks $m,m+1$ are possible, each of size $M$. Let $\mathcal B=\mathcal A_{m+1}$. Antichainness gives $\mathcal A_m\cap\partial\mathcal B=\varnothing$, and local LYM gives $|\partial\mathcal B|\ge|\mathcal B|$. Since the two <antichain> parts together have size $M$, equality must hold. The inclusion <graph> between these ranks is $(m+1)$-regular on each side. Its <edges> from $\mathcal B$ already number $(m+1)|\mathcal B|$, so equality leaves no edge from $\partial\mathcal B$ to the complementary upper <vertices>.
The <graph> is connected: exchanging one element between two $m$-sets joins them through their $(m+1)$-element union, and successive exchanges connect all lower <vertices>; every upper vertex has a lower neighbour. Consequently $\mathcal B$ is either empty or the entire upper rank. This <connectedness of adjacent-level incidence in a Boolean lattice> excludes mixed extremizers. Hence
$$
\boxed{\text{the maximum antichains are exactly the full middle levels}.}
$$
There is one choice for even $n$ and two for odd $n$. For $n=0$, the sole maximum family is $\{\varnothing\}$, agreeing with this description.
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