A graph with number of vertices less than R(n,n)1, is called quasiRamsey of case n, if it has no complete graph Kn in itself or its complementary graph, and if added another vertex, no matter how it is lined to the graph, the derived graph will have Kn in itself or its complementary. When the number of vertices is R(n,n)1, call it a Ramsey graph. There is a simple example to show the existence of quasiRamsey graph. First R(n,n)>(n1)(n1). Second, choose n1 (n1)complete graphs Kn1, without any lines between any two vertices from two distinct Kn1. This graph with (n1)(n1) vertices is a quasiRamsey graph. Furthermore, there are a series of questions about it. For example, if there could be nonisomorphic quasiRamsey graphs with same number of vertices, if they are all completely symmetrical (vertex transitive)and how many quasiRamsey graphs there are. If all the quasiRamsey graphs can be cleared out, we will get a deeper understanding of the structure of Ramsey graphs. Then we are more likely to obtain better estimates of Ramsey numbers.

In the case of $R(4,4)$, there is a unique graph $G$ of order 17 and two nonisomorphic graphs $G_1, G_2$ of order 16. You can find the graphs here. The three graphs $G,G_1,G_2$ are each isomorphic to their own complements. In the case of $R(5,5)$, the largest known graphs are on 42 vertices, even though the best proven upper bound on their size is 48. All of the 656 known graphs of order 42 are available on the same page, and the paper explaining their construction is B. D. McKay and S. P. Radziszowski, Subgraph counting identities and Ramsey numbers, J. Combinatorial Theory, Ser. B, 69 (1997) 193209 PDF. In general I don't think much is known about the Ramsey graphs slightly below the maximal graphs, and it is hopeless at the current state of knowledge to find general constructions for them. ADDED: Thanks for your reply, I didn't notice you wanted vertexmaximal Ramsey graphs. The second of the (4,4)graphs of order 16 is one. There are 634 of order 15, 126010 of order 14, 693270 of order 13, 136834 of order 12, 4124 of order 11, 76 of order 10, 6 of order 9, none of smaller order. The great majority of these are not regular. The 6 of order 9 include your example $3K_3$ (three triangles). 


The complement of a quasiRamsey graph will still be quasiRamsey so that answers the question about nonisomorphic with the same number of variables (together with your construction). $R(4,4)=18$ and there is a unique 17 vertex graph (a Paley graph) with no $K_4$ subgraph or 4 vertex independent set.It is self dual and regular of degree 8. It is vertex transitive which means that except for one 16 vertex graph every other either contains 4 independent vertices or a $K_4$ or is quasiRamsey. This should make it easy to find some irregular 16 vertex quasiRamsey graphs. later Thanks to the answer from Brendan McKay (who would know!) I can say a bit more. I'll allow myself an editorial moment to say that the description of the graph6 format is not the most welcoming to the casual (windows based) curious reader who just wants to see the (adjacency matrix of) a particular graph or two. Hence I will state that the data pages he links to give descriptions of certain graphs but I have seen them. An R(4,4) graph is one which has neither 4 vertices with all 6 edges between them nor 4 vertices with 0 edges between them. Of course any subgraph of such a graph is also R(4,4). The definition in the question of a QuasiRamsey graph (for $n=4$) is an R(4,4) graph (on 16 vertices or less) so that no extension by adding one vertex and some edges on it is also R(4,4) As Brendan says (with a link to a page with the descriptions) there is a single 17 vertex R(4,4) graph. It has only one 16 vertex subgraph . There is exactly one other 16 vertex R(4,4) graph so it fits the given description of QuasiRamsey. Between them these two graphs have no more that 32 vertex subgraphs (probably somewhat less.) In all there are evidently 640 15vertex R(4,4) graphs so at least 608 of them are QuasiRamsey. Between all 640 there at most 640*15=9600 14vertex subgraphs. Evidently there are 130816 14vertex R(4,4) graphs so at least 1308169600= 121216 of them are QuasiRamsey. Most likely some of those are not regular (let alone vertex transitive). All this leads me to suspect that there are a very very large number of QuasiRamsey graphs for each $n \ge 5$ and that plenty of them are not regular. Perhaps someone can give an explicit description of at least one for $n=4$, however I can not at the moment. 

