4. By the condition, $a_{k}=\frac{a_{1}+a_{2}+\ldots a_{k-1}+a_{k+1}+\ldots a_{n}}{k+1}=\frac{120-a_{k}}{k+1}$ for all $k=1,2, \ldots, n-1$. From this, $(k+2) a_{k}=120$ and $a_{k}=\frac{120}{k+2}$. Considering that $a_{k}$ are integers, the maximum value of $n$ is determined by the divisibility of the numerator by $3,4,5, \ldots, n,(n+1)$. Since 120 is divisible by $3,4,5,6$, but not by 7, we get $n+1=6$, hence $\quad n_{\max }=5$. We calculate $\quad a_{1}=\frac{120}{3}=40$, $a_{2}=\frac{120}{4}=30, a_{3}=\frac{120}{5}=24, a_{4}=\frac{120}{6}=20$. Now we find the remaining number $a_{5}=120-(40+30+24+20)=6$.
Answer: $n_{\max }=5, a_{1}=40, a_{2}=30, a_{3}=24, a_{4}=20, a_{5}=6$.