No, this is not possible.
Assume there exists a model $M$ of ZF which is also a model of $Con(ZF + \varphi)$. This means that there is no integer in $M$ that codes a proof of $\lnot \varphi$ from $ZF$. Since every integer in "our" universe has a corresponding integer in $M$, none of "our" integers can code such a proof. Therefore $Con(ZF + \varphi)$ is true.
(EDIT: Alternatively, $M$ has within it what it believes to be a model $N$ of $ZF + \varphi$. But $N$ must then really be a model of $ZF + \varphi$.)
Thus $Con(ZF + Con(ZF + \varphi))$ implies $Con(ZF + \varphi)$. Therefore the situation you're speaking of can't arise.
EDIT: The statement in Tetori's problem may need to be changed as follows in order for my arguments to be conclusive. Thanks to GME for pointing out the error. However, even this formulation turns out to be trivially true for another reason.
Assume that ZFC + Con(ZFC) is consistent. Can there be a sentence $\varphi$ such that both of the following are theorems of ZFC + Con(ZFC)?
- Con(ZF + Con(ZF + $\varphi$)).
- Con(ZF + Con(ZF) + $\lnot$Con(ZF + $\varphi$)).
Then the answer is no, at least for the reasons above, as well as the fact that for any model of ZF, there is a model of ZFC with the same integers.
Tetori's original question had the following instead of 2.
2'. Con(ZF + $\lnot$Con(ZF + $\varphi$)).
2' is a consequence of Con(ZF) by Gödel's incompleteness theorem, so in this case, the question reduces to whether 1 can ever be a theorem of ZFC + Con(ZFC). But I don't think it can! Perhaps someone else knows the answer to this. So even the question as I reformulated it probably isn't all that interesting.
I think it probably would have been appropriate to replace ZFC + Con(ZFC) in the original question with ZFC + Con(ZFC + Con(ZFC)). My head is beginning to hurt, so I won't say anything else about this for now, but I think this might fix things.