Solution during contest: The main idea is to take the polynomial $P(X)=\prod_{i=1}^n{(X^2+Xa_i+b_i)}$ and notice the degree is 2n and has 2n roots namely $a_1,a_2, \dots a_n,b_1,b_2, \dots b_n$ so the polynomial is actually $P(X)=\prod_{i=1}^n{(X-a_i)(X-b_i)}$. Suppose $n \ge 3$( the first 2 you get answers) $\newline$ Now the coefficient of $X^0$ is on the one hand $\prod_{i=1}^n{b_i}$, on the other hand is $\prod_{i=1}^n{a_i} \cdot \prod_{i=1}^n{b_i}$. Now if $\prod_{i=1}^n{a_i}=1$ then you easiliy get 2 numbers equal, contradiction. So $\prod_{i=1}^n{b_i}=0$ WLOG $b_1=0$ so the other $2n-1$ numbers are not $0$. Dividing by X the eqaution becomes $\newline$ $(X+a_1)(X^2+Xa_2+b_2) \dots (X^2+Xa_n+b_n)=(X-a_1)(X-a_2)\dots (X-a_n)(X-b_2)\dots (X-b_n)$. Now comparing the coefficient of $X^0$ again we get $\newline$ $a_2a_3 \dots a_n=-1$. $n \ge 4$ will produce 2 equal numbers, contradiction so n=3 and WLOG $a_2=1, a_3=-1$ then go back in the equation, solve for $b_i$ and get contradiction.