Obviously $a_2=a_1$. Also $a_3=max(a_1,a_2-a_1)=a_1$. We have $a_1=a_2=a_3$. We will proceed by using induction. Suppose that $a_1=a_2=\cdots=a_p$ holds, we will prove $a_{p+1}=a_p$. On the other hand, for all $k=1,2,\cdots,p-1$ $$f_p(k)=\dfrac{\sum_{i=k+1}^{p}{a_i}}{n-k}+\dfrac{\sum_{i=1}^{k}{a_i}}{k}=a_k-a_1=0$$Thus, we obtain the following $$a_{p+1}=max(f_p(0),f_p(1),\cdots,f_p(p-1))=max(a_1,0,0,\cdots,0)=a_1$$implying that the sequence $(a)_1^n$ must be eventually constant as desired.