There do not exist four such polynomials. We show this as follows. Suppose that there do exist four such polynomials. If a polynomial has no real roots, it is either positive for all real $x$, or else it is negative for all real $x$. Consider the complete graph with the four polynomials as vertices.

Colour the edge $P_iP_j$ white if $P_i(x) + P_j(x) > 0$ for all real $x$, and black if $P_i(x) + P_j(x) < 0$ for all real $x$. We cannot have a triangle $P_iP_jP_k$ of the same colour, because, if we did, the polynomial $P_i(x) + P_j(x) + P_k(x)$ which is equal to a half of $(P_i(x) + P_j(x)) + (P_j(x) + P_k(x)) + (P_k(x) + P_i(x))$ has a constant sign for all real $x$, and therefore this sum of three polynomials would not have a real root.

By the Pigeonhole Principle, at least three of the six edges must have the same colour – let's say this colour is black. If three black edges have a common vertex, then in order to avoid forming a black triangle, the other three edges would need to be white, thus forming a white triangle – a contradiction.

So, without loss of generality, we can consider the case where $P_1P_2, P_2P_3$ and $P_3P_4$ are black. Then $P_1P_3$ and $P_2P_4$ must be white and therefore we have $(P_1(x) + P_3(x)) + (P_2(x) + P_4(x)) > 0$ for all real $x$. But since $P_1P_2$ and $P_3P_4$ are black, we have $(P_1(x) + P_2(x)) + (P_3(x) + P_4(x)) < 0$ for all real $x$. The case where $P_1P_2, P_2P_3$ and $P_3P_4$ are white yields a similar contradiction.

We conclude that there do not exist four such polynomials.