Call a graph $G$ $n$-saturated if for every set $A$ of size $n$ of vertices and all $B\subseteq A$ there is a vertex $v\not\in A$ that forms an edge with all $w\in B$ and 
does not form an edge with any $w\in A\setminus B$. 


A countably infinite graph is isomorphic to the random graph iff it is $n$-saturated for all $n$.  

Here are the questions:


1. Is it true that for all $n$ there is a finite graph (of size at least $n$) that is $n$-saturated?
If yes, are there reasonable upper and lower bounds on the size of such a graph?  

(An $n$-saturated graph of size $\geq n$ has at least $n+2^n$ vertices, but this is probably far from optimal.)
2. Is every finite graph an induced subgraph of a finite $n$-saturated graph?
(Edit: As Ori Gurel-Gurevich pointed out, the second question is silly. Clearly, if $G$ is $n$-saturated (and of size at least $n$), then it has every graph with $n$-vertices as an induced subgraph.)


Example: There is a $2$-saturated graph: Let the vertices of $G$ be the $2$-element subsets of a set with $6$ elements. Two of those vertices form an edge if they (as sets) have a non-empty intersection. It is easily checked that this graph is $2$-saturated.
But it also has 30 vertices. That seems a lot.



Finite graphs that realize all types over $n$-element sets
1. If $G=G(m,\frac12$) then for any fixed set $A$ and subset $B$ and $v$ we have the probability that $v$ is connected to $B$ but not to $A\setminus B$ is $2^{-n}$. The probability that all $v$ fail to do so is $(1-2^{-n})^{m-n} \approx e^{m 2^-n}$. There are less than $(2m)^n$ such subsets $A$ and $B$, so union bound gives that when $m=C 2^n n^2 $ (for some large $C$) the probability that for some $A$ and $B$ there will be no witness $v$ is negligible.
2. I'm not sure I understood the question. If $G$ is $n$-saturated, then any graph of size $\le n$ is an induced subgraph.


