Answer: no.
Assume that there exist such four quadratic polynomials $f_1, f_2, f_3$ and $f_4$. If a quadratic polynomial $f(x)$ has no real root, then either $f(x) > 0$ for all $x \in \mathbb{R}$ (we write $f > 0$ and say $f$ is positive) or $f(x) < 0$ for all $x \in \mathbb{R}$ (we write $f < 0$ and say $f$ is negative).

Lemma. If $g_1, g_2, g_3$ are (quadratic) polynomials such that the sum of any two of them has no real root and the sum $g_1 + g_2 + g_3$ has a real root, then $g_1 + g_2, g_1 + g_3$ and $g_2 + g_3$ all cannot have the same sign.

Proof. If $g_1 + g_2, g_1 + g_3$ and $g_2 + g_3$ are all positive, then $g_1 + g_2 + g_3$ is also positive and thus has no real root, which is a contradiction. $\square$

By the lemma, the sums $f_1 + f_2, f_1 + f_3, f_2 + f_3$ cannot have the same sign. Without loss of generality we can assume that
$$\begin{aligned} f_1 + f_2 &> 0 \\ f_1 + f_3 &> 0 \\ f_2 + f_3 &< 0. \end{aligned}$$

There are two possibilities.

Case 1: If $f_2 + f_4 > 0$, then applying the lemma for $f_1, f_2, f_4$, we get $f_1 + f_4 < 0$. This gives the contradiction
$$0 > (f_2 + f_3) + (f_1 + f_4) = (f_1 + f_3) + (f_2 + f_4) > 0.$$

Case 2: If $f_2 + f_4 < 0$, then applying the lemma for $f_2, f_3, f_4$, we get $f_3 + f_4 > 0$. Now applying the lemma for $f_1, f_3, f_4$, we get $f_1 + f_4 < 0$. This gives the contradiction
$$0 > (f_2 + f_3) + (f_1 + f_4) = (f_1 + f_2) + (f_3 + f_4) > 0.$$
Hence no such polynomials exist.