any solutions?

First Solution: (by a friend) Let $P_i$ be the number of people in apartment $i$, we will show that after each fight the sum $\displaystyle \sum \limits _{i=1}^{120}P_i^2$ decreases by at least $2$. Since the sum is always positive, the process terminates. Assume WOLG that the people in apartment $1$ fight and move to apartments $2,\ldots ,P_1+1$, then the value of the sum after the fight is $\displaystyle 0^2+\sum \limits _{i=2}^{P_1+1}(P_i+1)^2+\sum \limits _{i=P_1+2}^{120}P_i^2$. Hence the difference is\begin{align*}\Delta & =\sum \limits _{i=2}^{P_1+1}(P_i+1)^2+\sum \limits _{i=P_1+2}^{120}P_i^2-\sum \limits _{i=1}^{120}P_i^2 \\ & =\sum \limits _{i=2}^{P_1+1}(2P_i+1)-P_1^2 \\ & =2\sum_{i=2}^{P_1+1}P_i+ P_1-P_1^2 \\ & <2(119-P_1)+P_1-P_1^2 \\ & =238-P_1-P_1^2 \\ & <238-15-15^2 \\ & =-2, \end{align*}as claimed. Second Solution: (by my brother) If an apartment has a fight at a night $N$, then it has at least $15$ persons in the night $N$. The apartment that fights the night $N+1$ has at least $14$ persons in the night $N$ (at most $1$ person moves into the apartment). The apartment that fights the night $N+2$ has at least $13$ persons in the night $N$ (at most $2$ persons move into the apartment). $\vdots$ The apartment that fights the night $N+14$ has at least $1$ person in the night $N$ (at most $14$ persons move into the apartment). Observe that the people and the apartments are all distinct because people only move when they fight and you can't overcrow an empty apartment in less than $15$ nights. If the process lasted at least $15$ nights we should have at least $1+2+\cdots +15=120$ persons. Hence the process terminates after $14$ nights (at most).