for the same question in a prior posting, add the following size-bounding condition, that is: any subset of $A$ that is equal in size to a proper initial segment of $A$, is bounded in $A$ (with respect to $<^A$). The same property is to be applied to $B$.
The same question is to be placed but with the above condition holding, that is:
What's the least order type of $A$?
I'm under the impression that this must be a large cardinal, but I'm not sure of that.
The same counting argument as in your other question gives a not-too-big upper bound - namely, $\aleph_{\omega_1+1}$ (remember that $\aleph_{\omega_1}$ isn't regular). Specifically, there are only countably many definable (in the sense of second-order logic, or indeed with respect to any fixed countable logic) subsets of $\aleph_{\omega_1+1}$ but uncountably many regular cardinals $<\aleph_{\omega_1+1}$, so there must be some regular cardinal $\kappa<\aleph_{\omega_1+1}$ such that $(\kappa,\aleph_{\omega_1+1})$ is a good pair.
Of course, in the context of cardinals $<\aleph_{\omega_1}$ this $\kappa$ must be extremely large, just as the smallest element of a good pair in general must be extremely large in the context of countable ordinals. (The "$<\aleph_{\omega_1}$" isn't a typo: since $\kappa$ is a regular cardinal, $\kappa<\aleph_{\omega_1+1}$ implies $\kappa<\aleph_{\omega_1}$.)