There are $2n$ people. Some subsets of people are called nice; the empty subset is nice. For any nice subset $A$ with $i<n$ people, there are at least $2(n-i)$ people $x\not\in A$ such that $A\cup\{x\}$ are also nice. At least how many nice subsets are there?
Some thoughts: Since $\emptyset$ is nice, all $2n$ one-person sets are nice. On the other hand, it is possible to make all sets of size $\leq n$ nice; this satisfies the problem condition. Asymptotics/other bounds are also welcome.
On
It’s always possible to find a family of exactly $3^n$ nice sets. Here’s a construction.
Let $A=[n]\times[2]$. For $S\subseteq A$ let $S_1=\{k\in[n]:\langle k,1\rangle\in S\}$ and $S_2=\{k\in[n]:\langle k,2\rangle\in S\}$. Let
$$\mathscr{N}=\{S\subseteq A:S_1\cap S_2=\varnothing\}\;.$$
If $S\in\mathscr{N}$, then $S\cup\{\langle k,i\rangle\}\in\mathscr{N}$ for each $\langle k,i\rangle\in\big([n]\setminus(S_1\cup S_2)\big)\times[2]$, and
$$\left|\big([n]\setminus(S_1\cup S_2)\big)\times[2]\right|=2(n-|S|)\;,$$
so $\mathscr{N}$ is a possible family of nice subsets of $A$.
For $0\le k\le n$ there are $\binom{n}k$ ways to choose $S_1\cup S_2$ and $2^k$ ways to distribute $k$ elements of $S_1\cup S_2$ between the two sets, so there are $\binom{n}k2^k$ members of $\mathscr{N}$ of cardinality $k$. Thus,
$$|\mathscr{N}|=\sum_{k=0}^n\binom{n}k2^k=3^n\;.$$
There must be at least $3^n$ nice sets.
Let $a_i$ be the number of nice sets of size $i$.
$a_0=1$ since $\emptyset$ is nice.
For $0\le i\lt n$ we have $\frac{a_{i+1}}{a_i}\ge\frac{2(n-i)}{i+1}$, since each nice set of size $i$ is contained in at least $2(n-i)$ nice sets of size $i+1$, while each nice set of size $i+1$ contains at most $i+1$ nice sets of size $i$.
It follows by induction on $i$ that $a_i\ge\binom ni2^i$ for $0\le i\le n$, whence the total number of nice sets is at least $\sum_{i=0}^na_i\ge\sum_{i=0}^n\binom ni2^i=3^n$.
This bound is attained at least when $n\le3$. For $n=3$, if $\{0,1,2,3,4,5\}$ is the set of people, we can take as our nice sets: the empty set; all six $1$-element sets; all $2$-element sets except $\{0,3\}$, $\{1,4\}$, $\{2,5\}$; the the eight $3$-element sets $\{0,1,2\}$, $\{1,2,3\}$, $\{2,3,4\}$, $\{3,4,5\}$, $\{4,5,0\}$, $\{5,0,1\}$, $\{0,2,4\}$, $\{1,3,5\}$.
P.S. The answer by Brian Scott shows how a family of $3^n$ nice sets can be constructed: namely, partition the set of $2n$ people into $n$ "nasty" pairs, and decree that a set is "nice" if and only if it contains no nasty pair as a subset. I will now show that, if the family of nice sets has cardinality $3^n$, then it must be of this form.
If there are $3^n$ nice sets, then the inequalities above must all be equalities. Thus the number of nice sets of size $i$ is exactly $\binom ni2^i$ for $0\le i\le n$, and so for $0\le i\lt n$ each nice $i$-set is contained in exactly $2(n-i)$ nice $(i+1)$-sets, and each nice $(i+1)$-set contains exactly $i+1$ nice $i$-sets; and of course there are no nice sets of size greater than $n$.
The empty set and all singletons are nice. Call a pair "nasty" if it's not nice. Since each nice singleton is contained in exactly $2n-2$ nice pairs, each nice singleton is contained in exactly one nasty pair; i.e., the nasty pairs constitute a partition of the set of $2n$ people. Since removing one element from a nice set always leaves a nice set, every subset of a nice set is nice. Hence, a nice set can contain no nasty pairs. Since there are just $3^n$ sets containing no nasty pairs, if there are $3^n$ nice sets, the collection of nice sets must coincide with the collection of all sets containing no nasty pairs.