In a question and an answer at MO, Joel David Hamkins showed that (if ZFC is consistent) there are models of ZFC in which $V\neq HOD$ and every $\Sigma\_2$-definable set has a definable member.


Let $\mathfrak{M}$ be such a model.
My question is: Can such a model $\mathfrak{M}$ satisfy further the following:


(\*) Every ordinal definable set of reals is Baire (or Lebesgue) measurable?


I suspect very much that the answer is negative, but I would like the confirmation of an expert. 



A question on models of set theory and Lebesgue measure
This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma\_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements. 


The question of whether a set of reals $A$ is measurable or not is something that can be checked in a comparatively small rank-initial segment of the universe, in $V\_{\omega+3}$ or so, and for this reason, it is a [local property](http://jdh.hamkins.org/local-properties-in-set-theory/), which therefore has complexity at worst $\Delta\_2$. 


