No, this cannot happen, although it's a little bit trickier than one might expect to prove this!

First, a miniature result:

Suppose $T,S$ are computably axiomatizable theories in the language of arithmetic, each containing the theory $\mathsf{I\Sigma_1}$, with $T\vdash Con(S)$ and $S\vdash Con(T)$. Then $T$ and $S$ are inconsistent.

*If you haven't seen $\mathsf{I\Sigma_1}$ before, the only points you need to know are that it is finitely axiomatizable, strong enough for Godel's theorems to be applicable, and ***self-provably** $\Sigma_1$-complete. Note that neither of the better-known arithmetics $\mathsf{Q}$ or $\mathsf{PA}$ will suffice: $\mathsf{Q}$ doesn't prove its own $\Sigma_1$-completeness since it lacks induction, and $\mathsf{PA}$ isn't finitely axiomatizable.

**PROOF**. It will be enough (by symmetry) to show that $T$ is inconsistent.

Since $\mathsf{I\Sigma_1}$ is finitely axiomatizable and proves its own $\Sigma_1$-completeness, we have that $T$ proves "$S$ is $\Sigma_1$-complete:" just verify in $T$ an $S$-proof of any single sentence axiomatizing $\mathsf{I\Sigma_1}$. Consequently, $T$ proves the sentence $\neg Con(T)\rightarrow [S\vdash (\neg Con(T))]$.

On the other hand, since $S\vdash Con(T)$ and $T$ is $\Sigma_1$-complete we have that $T$ proves $S\vdash Con(T)$. Putting this together with the above paragraph, we get a $T$-proof of "If $T$ is inconsistent, then $S$ proves $Con(T)\wedge\neg Con(T)$" - that is, a $T$-proof of $\neg Con(T)\rightarrow\neg Con(S)$.

But since $T\vdash Con(S)$, this gives a $T$-proof of $Con(T)$ - so $T$ is inconsistent.

The above can be improved, however.

First there's the issue of generalizing beyond $n=2$. This isn't very interesting though, since it's clear how to proceed: simply iterate the above idea by applying "provable $\Sigma_1$-completeness" over and over again.

More interestingly there's the *language* issue: $\mathsf{ZFC}$ for example is not a theory in the language of arithmetic, so the above result doesn't immediately apply to it. This can be handled via the notion of an **interpretation**. Basically, a theory $A$ interprets a theory $B$ if there is some tuple of formulas $\Phi_A$ in the language of $A$ such that for each sentence $\varphi\in B$, the theory $A$ proves that the structure defined by $\Phi_A$ satisfies $\varphi$. (Think about how $\mathsf{ZFC}$ implements arithmetic via the finite ordinals, for example.)

Via interpretations, we can generalize the argument above to arbitrary languages. Combined with the generalization past $n=2$ above, this gives the stronger result:

Suppose $T_1,...,T_n$ are computably axiomatizable theories, each of which interprets $I\Sigma_1$, such that $T_1\vdash Con(T_2)$, $T_2\vdash Con(T_3)$, ..., $T_n\vdash Con(T_1)$. Then each $T_i$ is inconsistent.

The most difficult part here is being precise about what "$Con(-)$" should mean in each of the relevant languages (basically, we just "work along interpretations").

The final improvement to be made is with respect to the base theory. We can replace $\mathsf{I\Sigma_1}$ with substantially weaker theories without changing the argument, but this doesn't get us all the way to $\mathsf{Q}$. So - dropping back to a more manageable level of generality along the other axes - we're left with a natural question:

Can there be two computably axiomatizable consistent theories $T,S$ in the language of arithmetic containing $\mathsf{Q}$ such that $T\vdash Con(S)$ and $S\vdash Con(T)$?

As Emil Jerabek comments below, the answer is **still** negative. However, at this point I'm not familiar with the relevant methods, so I can't say anything meaningful.