I like the answer of Michael Brown, but it has a "hole". There is a simple counter-argument to this answer: do not consider the growing crystal alone, but take it with the whole environment: crystal, vapor and melt altogether. Then one again comes to the same question of Kitchi.
The origin of the misunderstanding is just the use of entropy as of a thermodynamic potential. Theoretically any potential might be used for description. In practice, however, the entropy is very inconvenient, since one of its variables is the mean internal energy. It has the effect that in order to be able to discuss a process in terms of its entropy one should make sure that the mean energy is constant during this process. Which is not the case during crystallization. Indeed, as soon as a molecule joins the crystal from the melt, the energy goes down.
The problem is removed, if one uses for these purposes the free energy, or even better the omega-potential, rather than the entropy. In the terms of these two potentials it is easier to think about phase transitions. After all, we got used to these terms much more, since they are akin to mechanical energy: the system goes to the state with the smallest free energy possible. In these same state the entropy will exhibit the absolute maximum.
One can now ask himself, if the free energy goes down during crystallization? The answer is "Yes, of course". Since, if under some conditions it does not go down with crystallization, the crystallization will not take place: only processes in which the free energy decreases leads to the thermodynamic equilibrium state. Then, does the total system entropy in this process go up? The answer is "Yes, of course", since, where the free energy goes down, the entropy necessarily goes up.