Let $q_i$ be the $i$'th polynomial in the list in the previous post, and let $x_i$ be the root of $p_i$. Assume there are three $q$'s with no real roots. We may assume WLOG that $q_1,q_2,q_3$ have no real roots. Now let $P=\prod_{i\ne 1,2,3}p_i$. We thus know that $p_1p_2P+p_3,p_2p_3P+p_1,p_3p_1P+p_2$ have no real roots. By plugging the roots of $P$ in these polynomials, we find that $P(x_1),P(x_2),P(x_3)$ all have the same sign. Now plug $x_i$ in $q_i$ to get a contradiction: we find $(x_1-x_2)(x_1-x_3),(x_2-x_1)(x_2-x_3),(x_3-x_1)(x_3-x_2)$ to have the same sign, which is impossible.

I will assume that the coefficients of $p_i$ are real. Let us sort $p_i$ by the size of its root $r_i$. Let $q_i = \prod_{j = 1, j \neq i}^n p_j +p_i$. Clearly there is a sign change between $q_i(r_{i-1})$ and $q_i(r_{i+1})$ for all $2 \le i \le n-1$ and that is n-2 of the polynomials.