SAUDI ARABIAN MATHEMATICAL COMPETITIONS

The answer $n=3$. In fact, the polynomials $P_1(x)=x^2-4$, $P_2(x)=x^2-4x+6$ and $P_3(x)=x^2-8x+12$ satisfy the conditions:

$P_1+P_2=2(x-1{)}^2$,

$P_1+P_3=2(x-2{)}^2$,

$P_2+P_3=2(x-3{)}^2$.

Suppose that there are four polynomials $P_1,P_2,P_3,P_4$ satisfying the conditions. Then

$P_1+P_2=2(x-t_{12}{)}^2$,

$P_3+P_4=2(x-t_{34}{)}^2$,

$P_1+P_3=2(x-t_{13}{)}^2$,

$P_2+P_4=2(x-t_{24}{)}^2$,

where $t_{ij}$ is the multiple root of the polynomial $P_i+P_j$.

Let $Q=P_1+P_2+P_3+P_4$. Then $Q$ has two representations

$$Q=2(x-t_{12}{)}^2+2(x-t_{34}{)}^2\text{ and }Q=2(x-t_{13}{)}^2+2(x-t_{24}{)}^2.$$

By considering linear and constant terms of both expressions of $Q$, we get

$$t_{12}+t_{34}=t_{13}+t_{24}\text{ and }{t}_{12}^2+t_{34}^2=t_{13}^2+t_{24}^2.$$

This shows that $(t_{12},t_{34})=(t_{24},t_{13})$. But, if $t_{12}=t_{13}\Rightarrow P_2=P_3$, and if $t_{12}=t_{24}\Rightarrow P_1=P_4$, in both cases, we get a contradiction. $\square$