Question: I wanted to ask about the following problem:

An ideal monatomic gas is separated into two volumes $V\_{1}$ and $V\_{2}$ through a diathermic piston, such that each volume containing $N$ atoms and the two sides are at the same temperature $T\_{0}$. The entire system is isolated from the outside by means of insulating walls.

The plunger is reversibly externally manipulated until the two gases are in thermodynamic equilibrium with each other.

After the problem has different questions, however, I have not been able to answer the following:

How is called the type of process? Ie is isothermal, adiabatic, etc. Show that $\Delta S\_{1}$ = $-\Delta S\_{2}$, where $\Delta S\_{1}$ and $\Delta S\_{2}$ are the changes in entropy of the two gases.

Answer: 
We know that for reversible processes in isolated systems the entropy remains constant. You have mentioned that the system is isolated and that the plunger is reversibly manipulated. So considering the 2 gases as one isolated system we get that the entropy of the system is a constant. So $\Delta$$S$ = $0$
Now since there is a diathermic piston separating the gases, it means that one gas in itself is not an isolated system, because heat transfer can take place through diathermic piston. Thus considering each gas, there will be a non-zero change in entropy for each. So $\Delta$$S(1)$ + $\Delta$$S(2)$ = $\Delta$$S$ = $0$
Thus you get your result.
There can be heat interactions between the two volumes themselves if you treat them as two seperate systems. So even though the entire system is isolated heat interactions are possible between the two volumes.