= Solution
Proceed by induction on the ground-set size, starting with the one-element chain $\{\varnothing\}$ in the <Boolean lattice> on no coordinates. Suppose a <symmetric chain decomposition of a Boolean lattice> on $[n]$ has been constructed. Write one of its chains as
$$
S_r\subset S_{r+1}\subset\cdots\subset S_{n-r},\qquad |S_j|=j.
$$
On adjoining the new coordinate $v=n+1$, replace its two copies by
$$
S_r\subset\cdots\subset S_{n-r}\subset S_{n-r}\cup\{v\}
$$
and, when nonempty,
$$
S_r\cup\{v\}\subset S_{r+1}\cup\{v\}\subset\cdots\subset S_{n-r-1}\cup\{v\}.
$$
The first chain has endpoint ranks $r,n-r+1$, summing to $n+1$. The second has endpoint ranks $r+1,n-r$, also summing to $n+1$. Both are saturated chains, increasing rank by one at every step. If the original chain was a singleton, the second chain is omitted.
These new chains are disjoint: the first contains every old set without $v$ and just the last old set with $v$; the second contains precisely the remaining old sets with $v$. Thus together they cover both lifted copies of the old chain. Different old chains have disjoint copies, so performing this construction for each yields a partition of all subsets of $[n+1]$ into <symmetric chains>. This completes the induction and proves \b[a <symmetric chain> partition exists for every $n$].
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