1. The numbers $p$ and $b$ are roots of the quadratic equation $x^{2}+2020 a x+c=0, a \neq 0$. Find the sum of the roots of the quadratic equations $a x^{2}+b x+d=0$ and $a x^{2}+p x+q=0$, if each of them has 2 distinct real roots.
Solution. Since $p$ and $b$ are the roots of the quadratic equation $x^{2}+2020 a x+c=0$, by Vieta's theorem, $p+b=-2020 a$. Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2}+b x+d=0$, and $x_{3}$ and $x_{4}$ be the roots of the equation $a x^{2}+p x+q=0$. Then, by Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a}$ and $x_{3}+x_{4}=-\frac{p}{a}$. Therefore, $x_{1}+x_{2}+x_{3}+x_{4}=-\frac{b}{a}-\frac{p}{a}=-\frac{b+p}{a}=-\frac{-2020 a}{a}=2020$.

Answer: 2020.

Comment. The participant of the olympiad incorrectly refers to Vieta's theorem as the inverse theorem of Vieta - no points should be deducted.