Suppose I have a perfect crystal with statistical entropy $S_1$. If I cool it to 0 Kelvin, the number of possible microstates becomes 1, hence the statistical entropy becomes 0.
However, statistical entropy is only defined up to an additive constant. This is because there are actually infinitely many microstates, and you need to choose $\delta x$ and $\delta p$ in the course graining process.
Therefore, we could define the initial entropy in another way to see that it is a different number $S_2$. Now with this alternative choice of microstates, the entropy does still tends to zero when we let the temperature go to zero Kelvin.
How much entropy has left the system in the cooling process? Differences in entropy should be independent from the choice of microstates. However, it seems that $\Delta S$ is both $-S_1$ and $-S_2$, which is impossible?