Why does a null entropy variation along a closed timelike curve imply a reversible process? Carlo Rovelli in one of his articles from 2019 (reference: https://arxiv.org/abs/1912.04702) argues that time travel into the past are thermodynamically impossible:

For instance, if we want to travel to the past and arrive to the past keeping our memory of events happened in the future, we need some device (like our brain) capable of memory. But memory is an irreversible phenomenon (we remember the past not the future) and, like all irreversible phenomena, is based solely on the only fundamental irreversible law: the second principle of thermodynamics $\frac{dS(\tau)}{d\tau}\ge 0$. Along a closed timelike loop (CTC) $\gamma$ the only possibility of having $\frac{dS(\tau)}{d\tau}\ge 0$ everywhere is having $\frac{dS(\tau)}{d\tau}= 0$. But this means that all the processes around γ are reversible, and therefore there can be no memory.

Where $\tau$ parametrizes the loop $\gamma$.
Since S is a state function, we must have $\Delta S = 0$ along a CTC...However, if the change in entropy of the system that runs through the CTC is zero, it does not imply that the process is reversible ... this is the thing I don't understand.
Perhaps Rovelli means that the device that runs through the CTC performs a reversible transformation and irreversibility is "discharged" into the external environment?
I'm not sure about the latter explanation, any help is appreciated.
Addendum
If we consider a physical system along a CTC, then we can write:
$\Delta S_{Universe}=\Delta S_{surrounding}+\Delta S_{system}$;
$\Delta S_{system}=0$ for any process along the CTC, but $\Delta S_{surrounding}\ne 0$ in general, thus $\Delta S_{Universe}\ne 0$.
So, how can we say that any process is reversible along the CTC?
 A: In order to travel past, you need a device that records these events (since that's one of the way to tell that you have traveled to past). But recording is a process where entropy increases. That is $$dS>0$$
Meanwhile, the entropy is a state function. Thus for a CTC ($\gamma$) we have
$$dS=0$$
Hence, a device that records these events (like our brain), cannot exist in CTC (because if it was possible, it would be contradicting the 2nd Law of Thermodynamics)

Since S is a state function, we must have ΔS=0 along a CTC...However, if the change in entropy of the system that runs through the CTC is zero, it does not imply that the process is reversible ... this is the thing I don't understand.

Another possibility is that some parts of the curve say, $\gamma_+ = [0, \pi]$, we record these events and entropy of the device increases (an irreversible process, $dS>0$) But in the second part of the curve, say $\gamma_- = [\pi, 2\pi]$, the entropy of the device must decrease ($dS<0$).
However, while in $\gamma_-$, we cannot move forward in time and also see a decrease in the entropy of the device.
Edit:

If we consider a physical system along a CTC, then we can write: ΔSUniverse=ΔSsurrounding+ΔSsystem;


ΔSsystem=0 for any process along the CTC, but ΔSsurrounding≠0 in general, thus ΔSUniverse≠0.


So, how can we say that any process is reversible along the CTC?

As you have said "along the CTC", the process is reversible since its about the CTC.
In earth, I can create many systems that entropy stays the same (thus a reversible process) but the entropy of the universe increases. So and entropy increase in the universe does not mean that the system that I have created is irreversible.
And the entropy of the universe does not have to increase...even our universe will reach to a maximum entropy state at some point.
