The smallest such $n$ is $n=3$. For $n=2$, let $a_1 = \frac{a}{b}$, $a_2 = \frac{c}{d}$ where $\gcd(a,b) = \gcd(c,d) = 1$. Then we must have $bd \mid ad+bc$, $ac \mid ad+bc$. Taking mod $d$ in the first condition, we get that $d \mid bc$ but since $c,d$ are coprime we have that $d|b$. Taking mod $b$, we get $b \mid ad$ and since $a,b$ are coprime we get $b \mid d$ which tells us that $b=d$. Similarly we get $a=c$ and hence we have $a_1=a_2$. Clearly only $a_1=a_2=1$ or $\frac{1}{2}$ works and we only have finitely many such duplets.

Now for $n=3$, we will look for triplets of the form $(\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c})$ where $a,b,c$ are positive integers. Fixing $a=1$, it suffices to find infinitely many pairs of $(b,c)$ such that $\frac{1}{b}+\frac{1}{c}+\frac{b}{c}+\frac{c}{b}$ is an integer. We shall show that there are infinitely many solutions to $\frac{1}{b}+\frac{1}{c}+\frac{b}{c}+\frac{c}{b}=3$ by Vieta Jumping. If we let $c \ge b$, the equation rearranges into $b^2-b(3c-1)+c^2+c=0$. Viewing it as a quadratic in $b$, we get that there exists another $b'$ satisfying $\frac{1}{b'}+\frac{1}{c}+\frac{b'}{c}+\frac{c}{b'}=3$ and that this $b'$ satisfies $b'+b' = 3c-1$, $bb' = c^2+c$. Since $b$ is a positive integer, we get that $b' = 3c-1-b$ is also a positive integer. Furthermore, $b' = \frac{c^2+c}{b} \ge \frac{c^2+c}{c} > c$. Hence our triplet $(1,b,c)$ will transform into another triplet $(1,c,b')$ where $b' > c$ and we can then jump again but this time with $c$ as the subject of the quadratic. Starting with $b=3, c=6$, this algorithm will generate infinitely many distinct triplets of desired positive integers $(a,b,c)$ which will then give us infinitely many distinct triplets of rationals satisfying the problem conditions.