Do there exist four polynomials with real coefficients so that sum of every two doesn't have a real root while the sum of every three has a real root.
I have seen a similar problem with six polynomials which Ramsey's theorem could show that the answer is NO.
No, there are no such four polynomials.
We use two lemmas (the proof is written at the end of the answer) :
Lemma 1 : If $P(x)+Q(x)\gt 0, P(x)+R(x)\gt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.
Lemma 2 : If $P(x)+Q(x)\lt 0, P(x)+R(x)\lt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\gt 0$ for all $x\in\mathbb R$.
Suppose that four polynomials $P_1(x),P_2(x),P_3(x),P_4(x)$ satisfy our condition.
We may suppose that $$P_1(x)+P_2(x)\gt 0,\qquad P_1(x)+P_3(x)\gt 0$$ (At least two of $P_1(x)+P_2(x),P_1(x)+P_3(x),P_1(x)+P_4(x)$ have the same sign. We may suppose that $P_1(x)+P_2(x)$ and $P_1(x)+P_3(x)$ have the same sign. If both are negative, then we can name $-P_1(x),-P_2(x),-P_3(x),-P_4(x)$ as $P_1(x),P_2(x),P_3(x),P_4(x)$ respectively so that we have $P_1(x)+P_2(x)\gt 0$ and $P_1(x)+P_3(x)\gt 0$.)
From the lemma, we have $$P_2(x)+P_3(x)\lt 0$$
Suppose here that $P_1(x)+P_4(x)\gt 0$.
Then, since $P_1(x)+P_2(x)\gt 0$ and $P_1(x)+P_4(x)\gt 0$, it follows from the lemma that $P_2(x)+P_4(x)\lt 0$. Since $P_2(x)+P_3(x)\lt 0$ and $P_2(x)+P_4(x)\lt 0$, it follows from the lemma that $P_3(x)+P_4(x)\gt 0$. Since $P_1(x)+P_3(x)\gt 0$ and $P_3(x)+P_4(x)\gt 0$, it follows from the lemma that $P_1(x)+P_4(x)\lt 0$, which contradicts $P_1(x)+P_4(x)\gt 0$.
So, we have $P_1(x)+P_4(x)\lt 0$.
If $P_2(x)+P_4(x)\lt 0$, then since we have $P_2(x)+P_3(x)\lt 0$, it follows from the lemma that $P_3(x)+P_4(x)\gt 0$.
Now, we have two cases to consider :
Case 1 : $$P_1(x)+P_2(x)\gt 0,\quad P_1(x)+P_3(x)\gt 0,\quad P_2(x)+P_3(x)\lt 0$$ $$P_1(x)+P_4(x)\lt 0,\quad P_2(x)+P_4(x)\lt 0,\quad P_3(x)+P_4(x)\gt 0$$
In this case, we have $$P_1(x)\gt -P_2(x)\gt P_3(x)\gt -P_4(x)\gt P_1(x)$$ which is impossible.
Case 2 :
$$P_1(x)+P_2(x)\gt 0,\quad P_1(x)+P_3(x)\gt 0,\quad P_2(x)+P_3(x)\lt 0$$ $$P_1(x)+P_4(x)\lt 0,\quad P_2(x)+P_4(x)\gt 0$$ In this case, we have $$P_2(x)\lt -P_3(x)\lt P_1(x)\lt -P_4(x)\lt P_2(x)$$ which is impossible.
Finally, let us prove the lemmas.
Lemma 1 : If $P(x)+Q(x)\gt 0, P(x)+R(x)\gt 0$ for all $x\in\mathbb R$, and $Q(x)+R(x)$ has no real roots, and there exists a real number $\alpha$ such that $P(\alpha)+Q(\alpha)+R(\alpha)=0$, then $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.
Proof for lemma 1 :
Since $$Q(\alpha)=-(P(\alpha)+R(\alpha))\lt 0,\qquad R(\alpha)=-(P(\alpha)+Q(\alpha))\lt 0$$ we have $Q(\alpha)+R(\alpha)\lt 0$.
Suppose here that there exists a real number $\beta$ such that $Q(\beta)+R(\beta)\gt 0$.
Then, by the intermediate value theorem, $Q(x)+R(x)$ has at least one real root, which contradicts that $Q(x)+R(x)$ has no real roots.
It follows from this that $Q(x)+R(x)\lt 0$ for all $x\in\mathbb R$.$\qquad\blacksquare$
Lemma 2 can be proven in the similar way as above.
