Nice problem! I found the following solution, hope it's correct: Let $t$ be common root of some pair of trinomials $x^2+ax+b$ and $x^2+cx+d$, i.e \[ t^2+at+b=0=t^2+ct+d \Rightarrow t=\frac{d-b}{a-c} \in \mathbb{Q} \Rightarrow t \in \mathbb{Z} \]since leading coefficients of trinomials is equals to $1$. Also note that $|t| \leqslant N$ and $t < 0$ (since $x^2+ax+b$ doesn't has positive roots). Now we will prove the following Lemma: if $t$ is a integer number such that $-N \leqslant t \leqslant -1$, then there are at most $|N/t|$ ordered pairs $(a,b)$ of positive integers no exceeding $N$ such that $t^2+at+b=0$. Proof: let $k=-t$, so $k^2-ak+b=0$. Note that it implies $k|b$ and so $b=k \ell$ for some $\ell \in \mathbb{Z}_+$ and $\ell \leqslant N/k$, because $N \geqslant b=k \ell$. Then $k-a+\ell=0 \Rightarrow a=k + \ell$, so there are at most $N/k$ possible variations for $a$; note that $b$ is uniquely determinate knowing $a$. So, lemma is proved. $\ \square$. Now we are fully ready to solve the problem. Knowing that all common roots are minus positive integers numbers no more than $N$, and using lemma, we claim that pairs of trinomials having a common root not more than \[ \sum_{k=1}^{n} \binom{N/k}{2} < \frac{1}{2} \sum_{k=1}^{n} \left( \frac{N}{k} \right)^2 = \frac{1}{2}N^2 \left( \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots \right)< \frac{1}{2}N^2 \cdot 2 = N^2.\]Problem is solved!