For which smallest $n$ does there exist natural numbers $a_1,\dots,a_n$ making $\dfrac{(a_1+\cdots+a_n)^2-1}{a_1^2+\cdots+a_n^2}$ an integer?

Recently, at an olympiad I came across a problem that left me stuck.

Let $n\ge 2$ be a natural number. For which smallest $n$ do there exist natural numbers $a_1,a_2,\dots,a_n$ such that

$$ \frac{\bigl(a_1+a_2+\cdots+a_n\bigr)^2-1}{a_1^2+a_2^2+\cdots+a_n^2} $$

is an integer?

I am not very well-versed in number theory, but here are the attempts I made:

1. The case $n=2$.

   $$ \frac{(a+b)^2-1}{a^2+b^2}=\frac{2ab-1}{a^2+b^2}+1. $$

   Since $2ab-1<a^2+b^2$, the fractional part $\dfrac{2ab-1}{a^2+b^2}$ cannot be a nonzero integer. (Also $2ab-1=0$ is impossible in natural numbers.) Hence the expression cannot be an integer for $n=2$.

2. A general estimate using the Cauchy–Schwarz / inequality of arithmetic and quadratic means:

   $$ \bigl(a_1+\cdots+a_n\bigr)^2\le n\bigl(a_1^2+\cdots+a_n^2\bigr), $$

   so

   $$ \frac{\bigl(a_1+\cdots+a_n\bigr)^2-1}{a_1^2+\cdots+a_n^2}\le\frac{n\bigl(a_1^2+\cdots+a_n^2\bigr)-1}{a_1^2+\cdots+a_n^2}<n. $$

   Thus any integer value of the expression must be strictly less than $n$.

3. A parity observation. Since

   $$ a_1^2+\cdots+a_n^2\equiv\bigl(a_1+\cdots+a_n\bigr)^2\pmod 2, $$

   if the sum of squares is even then $\bigl(a_1+\cdots+a_n\bigr)^2-1$ is odd; therefore $a_1^2+\cdots+a_n^2\equiv a_1+\cdots+a_n\equiv 1\pmod 2$. (I tried to use parity to restrict possible values, but didn’t reach a useful contradiction.)

4. Trying to reduce to a Diophantine equation:

   $$ \bigl(a_1+\cdots+a_n-1\bigr)\bigl(a_1+\cdots+a_n+1\bigr)=k\cdot\bigl(a_1^2+\cdots+a_n^2\bigr), $$

   but that didn’t lead anywhere productive.

5. The equal-values test: take $a_1=\cdots=a_n=m$. Then

   $$ \frac{(nm)^2-1}{nm^2}=n-\frac{1}{nm^2}, $$

   which is not an integer for $nm^2\ne1$.

I would be grateful for any ideas or hints. Thanks for your interest in the problem!