The following theorem seems to express how the various
definability witness properties are connected with each other and
with $V=\text{HOD}$.

**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 $(\Sigma_2\wedge\Pi_2)$-definable nonempty set in $M$ has
an ordinal-definable element.

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

Proof. ($1\to 2$) The usual HOD order is a definable well-ordering
of the universe.

($2\to 3$) Select the least element with respect to the definable
order, as in Bjorn's answer.

($3\to 4$) Immediate.

($4\to 5$) Immediate.

($4\to 1$) If $M$ thinks there is a non-OD set, then the set $A$
of all non-OD sets in $M$ of minimal rank is a definable nonempty
set in $M$ with no ordinal-definable elements.

($5\to 1$) For the stronger implication — thanks to François
in the comments! — we analyze the complexity of the
definition of $A$ in the previous case. The main observation is
that $x\in\text{OD}$ is $\Sigma_2$, as every instance of
definability is witnessed by reflection inside some $V_\theta$.
The $\Sigma_2$ assertions are precisely the [semi-local
properties](http://jdh.hamkins.org/local-properties-in-set-theory/), those that can be
verified in some rank initial segment of the universe $V_\theta$.

If $V\neq\text{HOD}$, then the set $A$ of minimal-rank non-OD sets
is characterized by the following properties: $A$ is not empty;
all elements of $A$ have the same rank; every element of $A$ is
not in OD; every set of rank less than an element of $A$ is in OD;
every set not in $A$, but of the same rank as an element of $A$,
is in OD. Each of these properties is either $\Sigma_2$ or
$\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable.
Specifically, the first two requirements are $\Sigma_2$, being
witnessed in a rank-initial segment of the universe; the third is
$\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are
true just in case there is a large $V_\theta$ which believes them
to be true. Thus, the definition of $A$ has complexity
$\Sigma_2\wedge\Pi_2$, which allows the implication ($5\to 1$),
since $A$ has no ordinal-definable members.

($1\to 6$) Immediate, since under statement $1$, every set in $M$ is
ordinal-definable in $M$.

($6\to 4$) Immediate. QED

**Conclusion.** 
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.