I think this theorem expresses what is going on:

**Theorem.** The following are equivalent in any model $M$ of ZF:

1. $M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.

2. $M$ has a definable well-ordering of the universe.

3. Every definable nonempty set in $M$ has a definable element.

4. Every definable nonempty set in $M$ has an ordinal-definable element. 

5. Every ordinal-definable nonempty set in $M$ has an ordinal-definable element.

Proof. Statement 1 implies statement 2, using the usual HOD order. Statement 2 implies statement 3, by selecting the least element with respect to the definable order, as in Bjorn's answer. Statement 3 implies statement 4 trivially. Statement 4 implies statement 1, since if $M$ thinks there is a non-OD set, then the set of all non-OD sets in $M$ of minimal rank is a definable nonempty set in $M$ with no ordinal-definable elements. Statement 1 implies 5 since under 1 every set in $M$ is ordinal-definable in $M$. And statement 5 implies statement 4 trivially. QED

Thus, case (1) of the question occurs in exactly the models of $V=\text{HOD}$ that are not pointwise definable. There are such models, if ZFC is consistent, since one may take any uncountable model of $\text{ZFC}+V=\text{HOD}$.

Meanwhile, case (2) of the question does not occur at all, since if a set has sets that are not ordinal-definable, then it will have a definable set with no ordinal-definable members, namely, the set of all non-OD sets of minimal rank, as in the implication of statement 4 to statement 1.

Note that one may refine the theorem by paying attention to the complexity of the definition of the minimal-rank non-OD sets, which is $\Sigma_k$ for some very small finite number $k$ (probably $k\leq 5$, but I'd have to think about it). Thus, we could add statement 6, asserting that every nonempty $\Sigma_5$ set contains an ordinal-definable element, and this would also be equivalent to $V=\text{HOD}$.