Overpopulated apartment $119$ people live in a building with $120$ apartments. An apartment is called overpopulated if there are more than $15$ people living in it. Each day the tenants of some overpopulated apartment have a quarrel and they all move to different apartments in the same building. Is it true that eventually none of the apartments will be overpopulated?
Attempts and observations:

*

*If we organize the building into a $15\times 8$ grid with $8$ floors, after a quarrel at least $2$ of new floors(floors other than the floor on which the quarrel happened) occupy $1$ more tenants each(floors might be same)


*If an apartment is called overpopulated if there are more than $14$ people in it, then the answer is no


*If we prove that this process terminate, then we are done
 A: To try and prove that it is false would entail beginning from an overpopulated apartment, dispersing the inhabitants to different apartments  and getting another overpopulated apartment, and thus dispersing the inhabitants can cause the population of an apartment to increase by one for each abandoned apartment.
Say, you wanted a sequence of overpopulated apartments with $n_i$ people living in each one,
$$ n_i \geq 16 ,1 \leq i \leq m+1.$$
Beginning from $n_1$, $n_1$ people are dispersed to $n_1$ different apartments, if you want an apartment next with $n_2$ people in it, you would of needed to have an apartment with at least $n_2 -1$ people in it at first go. And then an apartment with at least $n_3 -2$ people in it at first go... and so on and so on.
Thus in general there must be $n_1 + n_2 -1 + n_3 -2 + ... =\sum\limits_{i=1}^{m+1} n_i - \sum\limits_{i=0}^{m}i \geq 16(m+1)-\frac{m(m+1)}{2}= 136> 119$ for $m=16$, since an apartment cannot have negative people in it. Thus if you wanted a sequence other than $16,16,16,...$ you would need even more people.
