Clearly, square of an integer number can give a remainder of $0$, $1$ or $4$ modulo $8$. Therefore, only for $n$ of the form $8k + r$, where $r \in \{0, 1, 4\}$, such numbers can exist. Let us show how they can be constructed.

$$n = 4t,\ a_1 = \dots = a_{n-1} = 1,\ a_n = (2t-1):$$
$$a_1^2 + a_2^2 + \dots + a_n^2 = (4t-1) \cdot 1^2 + (2t-1)^2 = (2t)^2.$$

$$n = 8t + 1,\ a_1 = \dots = a_{n-1} = 1,\ a_n = (2t-1):$$
$$a_1^2 + a_2^2 + \dots + a_n^2 = (8t) \cdot 1^2 + (2t-1)^2 = (2t+1)^2.$$