Let $a=a_1a_2,b=-2a_1a_2+na_1+a_2=-(a_1(a_2-n)+a_2(a_1-1)),c=a_1a_2-a_2-na_1+n=(a_1-1)(a_2-n)$ $a+b+c=n$ and $ax^2+bx+c=(a_1x-(a_1-1))(a_2x-(a_2-n))$ has roots $1-\frac{1}{a_1},1-\frac{n}{a_2}$

Can you explain what was the thought process there?

Let $q\in\mathbb{Z},$ $\lvert q\lvert\neq 1$. The equation $n[(q^2-q)x^2+(1-2q^2)x+(q^2+q)]=0$ has roots $\frac{q}{q-1}$ and $\frac{q+1}{q}$. It clearly gives infinite solutions for $n$.

If $ax^2+bx+c=0$ has a rational root then so does the quadratic equation $kax^2+kbx+kc=0$ for obvious reasons. So if we are able to prove the case when $n=1$, we can easily generate infintely many triples of $(a, b, c)$ in the case $n=k$ simply by letting $(ka, kb, kc)$. Rest is easy. Just let $(a, b, c)=(mn, -2mn+m+n, mn-m-n+1)$.