Problem 1820

Let's count the number of ordered pairs $(f, D)$, where $f$ is a labeling of vertices and $D$ a possible set of differences of labels of vertices joined by an edge.
Note that an ordered pair $(f, D)$ determines at most one great graph, since the labeling $f$ determine the vertices and the set of differences $D$ determine the edges of the graph (we join two vertices if and only if the difference between their labels is in $D$), unless there is some contradiction among the differences. In the latter case $(f, D)$ determines no great graph.
There are $\frac{\left\lfloor \frac{n^2}{4} \right\rfloor!}{\left(\left\lfloor \frac{n^2}{4} \right\rfloor - n\right)!} \le \left(\frac{n^2}{4}\right)^n$ labelings and since $D \subset \{1, 2, \dots, \left\lfloor \frac{n^2}{4} \right\rfloor\}$, there are $2^{\left\lfloor \frac{n^2}{4} \right\rfloor} \le 2^{\frac{n^2}{4}}$ possibilities for $D$. Thus the number of ordered pairs (and, henceforth, the number of great graphs with $n$ vertices) is not greater than $\left(\frac{n^2}{4}\right)^n \cdot 2^{\frac{n^2}{4}}$.
But the number of graphs with $n$ vertices is $2^{\binom{n}{2}} = 2^{\frac{n(n-1)}{2}}$. Therefore it suffices to show that for sufficiently large $n$
$$\left(\frac{n^2}{4}\right)^n \cdot 2^{\frac{n^2}{4}} < 2^{\frac{n(n-1)}{2}}$$
Taking logarithms on both sides,
$$n \cdot \log_2 \frac{n^2}{4} + \frac{n^2}{4} < \frac{n^2}{2} - \frac{n}{2} \iff 2 \log_2 \left(\frac{n}{2}\right) < \frac{n}{4} - \frac{1}{2}$$
which is true for sufficiently large $n$, since log grows slower than any polynomial with degree at least 1 and positive leader coefficient (it is true, indeed, for $n > 33$).