IRL_ABooklet

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 $$a_1=4,a_2=1,a_j=2(3\le j\le n-1),a_n=1.$$ It is easy to verify that this works. Many other constructions are possible.