Only $n = 2$ and $n = 3$ are interesting.

Note that if $n$ is not interesting, then any integer greater than $n$ is not interesting as well. So we just need to prove that $n = 4$ is not interesting.

We claim that there is no such example for the case
$$P_2(x) > P_4(x) > P_1(x) > P_3(x)$$
for infinitesimal positive number $x$.

For the sake of contradiction, assume that there exists an example. Since $P_i(x) - P_1(x)$ also satisfies the inequality, assume that $P_1(x) = 0$. Then by contradiction, every $P_i(x)$ has constant term zero.

Now, $P_3(x)$ is always negative for values close to $0$. Therefore, the term with the minimal degree in $P_3(x)$ is of the form $-a x^{2b}$, for some positive real number $a$ and positive integer $b$. Analogously, for $P_2(x), P_4(x)$, the terms of the minimal degrees are of the form $a x^{2b-1}$, for some positive real number $a$ and positive integer $b$. Let
$$\begin{align*} P_2(x) &= a_2 x^{2b_2-1} + \dots \\ P_3(x) &= -a_3 x^{2b_3} + \dots \\ P_4(x) &= a_4 x^{2b_4-1} + \dots \end{align*}$$
where every $a_i$ is a positive real number and every $b_j$ is a positive integer. Since $P_2(x) > P_3(x) > P_4(x)$ for $x < 0$, therefore
$$2b_2 - 1 > 2b_3 > 2b_4 - 1 \implies b_2 > b_3 > b_4 \implies P_2(x) < P_4(x),$$
for all sufficiently small positive real numbers $x$, contradicting the assumption $P_2(x) > P_4(x) > 0 > P_3(x)$.

But we claimed that $n = 2, 3$ are interesting. The case $n = 2$ is trivial and for $n = 3$, we have
$$P_i(x) = x Q_i(x) + c.$$
If $\sigma(i) = i$, we need to find polynomials $Q_i$ such that $Q_1(x) < Q_2(x) < Q_3(x)$ for $x \in (-\epsilon, 0)$ and $Q_1(x) > Q_2(x) > Q_3(x)$ for $x \in (0, \epsilon)$. It's clear that linear polynomials will work.

If $\sigma(1) = 1, \sigma(2) = 3$ and $\sigma(3) = 2$, let $Q_1(x) = 0$ and $Q_2(x), Q_3(x)$ be linear polynomials passing through point $0$ with negative leading coefficient. Other permutations are the same and we're done. ■