Show by induction that $a_n = (n+3)(n+2)(n+1)n$ for all $n \ge 3$.

First solution. The number $a_3 = 5 \cdot 72 = 360$ is an integer. For $n \ge 4$ the following equalities hold:
$$a_n = 5 \cdot \frac{a_1 + \dots + a_{n-1}}{n-1} \quad \text{and} \quad a_{n-1} = 5 \cdot \frac{a_1 + \dots + a_{n-2}}{n-2},$$
from which it follows that
$$a_n = \frac{5}{n-1} \left( \frac{n-2}{5} a_{n-1} + a_{n-1} \right) = \frac{n+3}{n-1} a_{n-1}.$$

$$a_n = \frac{(n+3)(n+2)\dots\cdot 7}{(n-1)(n-2)\dots\cdot 3} \cdot a_3 = \frac{(n+3)(n+2)(n+1)n}{6 \cdot 5 \cdot 4 \cdot 3} \cdot 360 = (n+3)(n+2)(n+1)n$$
is an integer, as required.

Second solution. Let $S_n = a_1 + a_2 + \dots + a_n$, then $a_{n+1} = S_{n+1} - S_n$.
To solve the problem, it is enough to show that all numbers $S_n$ are integers. From the condition we have $S_1 = 1$, $S_2 = 144$, and the formula $a_{n+1} = \frac{a_1 + a_2 + \dots + a_n}{n}$ takes the form $S_{n+1} - S_n = \frac{5S_n}{n}$, whence $S_{n+1} = \frac{n+5}{n} S_n$. Thus, for $n \ge 2$ we get

$$S_{n+1} = \frac{(n+5)(n+4)\dots\cdot 7}{n(n-1)\dots\cdot 2} S_2 =$$

$$= \frac{(n+5)(n+4)(n+3)(n+2)(n+1)}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \cdot 144 = \frac{(n+5)(n+4)(n+3)(n+2)(n+1)}{5}$$

Since at least one of the numbers $n+5, n+4, n+3, n+2, n+1$ is divisible by $5$, for $n \ge 2$ the number $S_{n+1}$ is an integer.