Suppose $(a_1, a_2, \dots, a_n)$ is an integer group which satisfies the conditions. Then by Cauchy's inequality we have
$$a_1^2 + \dots + a_n^2 \ge \frac{1}{n} (a_1 + \dots + a_n)^2 \ge n^3. \qquad \textcircled{1}$$
Combining $a_1^2 + \dots + a_n^2 \le n^3 + 1$, we see that it can only be $a_1^2 + \dots + a_n^2 = n^3$ or $a_1^2 + \dots + a_n^2 = n^3 + 1$.

If it is the former, then by the condition for Cauchy's inequality to take the equality sign, we can obtain $a_1 = \dots = a_n$. This requires $a_1^2 = n^2$, $1 \le i \le n$. Combining $a_1 + \dots + a_n \ge n^2$, we have $a_1 = \dots = a_n = n$.

If it is the latter, then let $b_i = a_i - n$, then we have
$$b_1^2 + b_2^2 + \dots + b_n^2 = \sum_{i=1}^{n} a_i^2 - 2n \sum_{i=1}^{n} a_i + n^3 \\ = 2n^3 + 1 - 2n \sum_{i=1}^{n} a_i \le 1.$$
Thus $b_1^2$ can only be $0$ or $1$, and there is at the most one among $b_1^2$, $b_2^2$, ..., $b_n^2$ to be $1$. If $b_1^2$, ..., $b_n^2$ all are zero, then $a_i = n$, $\sum_{i=1}^{n} a_i^2 = n^3 \ne n^3 + 1$. It leads to a contradiction. If there is just one among $b_1^2$, ..., $b_n^2$ to be $1$, then $\sum_{i=1}^{n} a_i^2 = n^3 \pm 2n + 1 \ne n^3 + 1$. Again, it leads to a contradiction.

Consequently, we obtain that it can only be $(a_1, \dots, a_n) = (n, \dots, n)$.