XIV OBM

Let $A$ have $n$ elements. The number of sequences of chains $$S_1\subset S_2\subset \cdots \subset S_{n-1}\subset A$$ such that $|S_i|=i$ is $n!$: for each permutation $(a_1,a_2,\dots ,a_n)$ of elements from $A$ let $S_i=S_{i-1}\cup \{a_i\}$.

Fix a subset $B$ of $A$. If $|B|=k$, then the chains containing $B$ $$S_1\subset S_2\subset \cdots \subset B\subset \dots S_{n-1}\subset A$$ is $k!(n-k)!$, because there are $k!$ choices for $S_1,\dots ,S_{k-1}$ and $(n-k)!$ choices for $S_{k+1},\dots ,S_{n-1}$. If $B$ and $B^{\prime }$ are such that $B\not\subset B^{\prime }$ and $B^{\prime }\not\subset B$, then every pair of chains, one containing $B$ and other containing $B^{\prime }$, are disjoint. So if $B_1,B_2,\dots ,B_m$ are subsets such that $B_i\not\subset B_j$ and $B_j\not\subset B_i$ then, denoting $|B_i|=m_i$, $$\sum _{i=1}^m{m}_i!(n-m_i)!\le n!⟺\sum _{i=1}^n\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{m_i}\right)}\le 1$$ The maximum value of $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ is $\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$. Hence $$\sum _{i=1}^m\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{m_i}\right)}\le 1⟹\sum _{i=1}^m\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)}\le 1⟺m\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$$

One example with $m=\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$ is considering all subsets with $\lfloor \frac{n}{2}\rfloor$ elements.