The construction for $n=9$ can be achieved with the polynomials $x^{2}+x+1, x^{2}+x+2$, and $x^{2}+5$. First we consider what kinds of polynomials we can have. Let $p(x)=(x+h)^{2}+k$. $h$ is either an integer or half an integer. Let $k=0$. If $h$ is an integer then $p(x)$ hits the perfect squares $0,1,4,9$, etc. If $h$ is half an integer, then let $k=1 / 4$. Then $p(x)$ hits the product of two consecutive integers, i.e. 0, $2,6,12$, etc. Assume there is a construction for $n=10$. In both of the cases above, the most a polynomial can hit out of 10 is 4, in the $0,1,4,9$ case. Thus $p_{1}$ must hit $1,2,5,10$, and $p_{2}$ and $p_{3}$ hit 3 integers each, out of $3,4,6,7,8,9$. The only ways we can hit 3 out of 7 consecutive integers is with the sequences $0,2,6$ or $0,1,4$. The only way a $0,2,6$ works is if it hits 3,5, and 9, which doesn't work since 5 was hit by $p_{2}$. Otherwise, $p_{2}$ is $0,1,4$, which doesn't work as $p_{2}$ hits 3,4, and 7, and $p_{3}$ must hit 6,8, and 9, which is impossible. Thus no construction for $n=10$ exists.