When $n = 3$, we can take the constant polynomials $f, g, h = -1, -2, 3$ which clearly satisfy the problem conditions.

Now assume that $n = 4$ and let our polynomials be $f_1, f_2, f_3, f_4$.
Note that for any $i, j$, we must have either $f_i(x) + f_j(x) > 0$ or $f_i(x) + f_j(x) < 0$ for all $x$ as otherwise it must have a real root. If there exist indices $i, j, k$ such that $f_i(0) + f_j(0), f_i(0) + f_k(0), f_j(0) + f_k(0)$ all have the same sign, say positive, then for all $x$
$$f_i(x) + f_j(x) > 0, \quad f_i(x) + f_k(x) > 0, \quad f_j(x) + f_k(x) > 0$$
$$\therefore f_i(x) + f_j(x) + f_k(x) > 0.$$
which is a contradiction as the sum $f_i(x) + f_j(x) + f_k(x)$ would have no real roots. We shall show that such a triple of indices must exist.

WLOG let $|f_1(0)| \ge |f_i(0)|$ for $i = 2, 3, 4$ and that $f_1(0) > 0$. Then $f_1(0) + f_i(0) > 0$ for all $i$. If there exist $i, j$ chosen from 2, 3, 4 such that $f_i(0) + f_j(0) > 0$, then $1, i, j$ is such a triple. If not, then 2, 3, 4 is such a triple. Thus $n$ cannot be four. Therefore the only answer is $n = 3$.