There is no positive solution for $n = 1$ or $n = 2$ but positive solutions exist for any $n \ge 3$. Eliminate the small cases first. If $n = 1$ then $1 = a_1 = 2$, a contradiction. With $n = 2$, we must solve $a_1 + a_2 = 4$ and $a_1 + 2a_2 = 4$. The unique solution is $a_1 = 4$ and $a_2 = 0$, and as $a_2$ is not positive, this is not feasible.

For $n \ge 3$, there are many ways to construct solutions, see Remark 2 below. Exploring heuristically, we can see that putting all $a_j = 2$ solves the first equation exactly but gives $n(n + 1)$ for the second equation, which is too high. We tweak that configuration, taking 1 from $a_2$ and from $a_n$, loading them instead onto $a_1$, which preserves the first equation and now satisfies the second: $(a_1, \dots, a_n) = (4, 1, 2, 2, \dots, 2, 2, 1)$ or more formally
$$a_1 = 4, \quad a_2 = 1, \quad a_j = 2 \ (3 \le j \le n-1), \quad a_n = 1.$$
It is easy to verify that this works. Many other constructions are possible.