Let $1 < a_1<a_2<\cdots < a_n$ be an increasing positive sequence of integers. Prove that there exists $a_{n+1} > a_n$ with $(a_1+\cdots + a_{n+1} )| (a_1^2 +\cdots + a_{n+1}^2)$.
One fact that could come in handy is that if $a,b$ are coprime then $a,b | c\Rightarrow ab | c.$
I think a proof by contradiction or the extremal principle could be useful. Suppose no such $a_{n+1}$ exists and let $(a_1,\cdots, a_n)$ be a strictly increasing sequence of positive integers with $a_1 > 1$ so that one cannot choose $a_{n+1} > a_n$ satisfying $(a_1+\cdots + a_{n+1} )| (a_1^2 +\cdots + a_{n+1}^2)$ and so that $a_n$ is minimal. The goal is to, if possible, construct another such sequence with $a_{n}$ strictly smaller than the previous one, giving the desired contradiction.
If the above approach doesn't work, perhaps an inductive proof might work? For the base case, consider $n=1$. Note that $a_2 = (a_1 - 1)a_1$ works for $a_1 > 2$, since have for $k=a_1 - 1$ that $a_1 + a_2 = (1+k)a_1$ and $(1+k) a_1 = a_1^2 | ((1+k^2)a_1^2).$ For $a_1 = 2,$ we can simply take $a_2 = 6.$ Now assume the result holds for some $n\ge 1$. We need to show the result holds for $n+1$.
For short, let us write $s:=a_1+\cdots+a_n$, $t:=a_1^2+\cdots+a_n^2$, and $x:=a_{n+1}$. The task is, given $s$ and $t$, find (suitably large) $x$ such that $x+s$ divides $x^2+t$.
We have $x^2+t=(x+s)(x-s) +s^2+t$. It will therefore be sufficient to find $x$ such that $x+s$ divides $s^2 +t$. Clearly $x=s^2+t-s$ will do. ($x>a_n$ because $x-s=(s-1)^2+t-1>0$ and $s\geqslant a_n$.)