Assuming the second condition is $P_1(x)>P_2(x)>\ldots>P_n(x)$ for $x\in (-\epsilon, 0)$, then the only interesting numbers are $n=2,3$. Note that if $n$ is not interesting then any integer greater than $n$ is not interesting, too. Lemma. If for two polynomials $H_i, H_j$ holds $H_i>H_j$ for $x\in(-\epsilon, 0)$ and $H_i<H_j$ for $x\in (0,\epsilon)$ then it must hold that $H_i(0)=H_j(0)$.(the same holds if the signs are reversed from $<$ to $>$ and vice versa) Proof. The polynomial $H_i(x)-H_j(x)$ is continuous and changes signs in $(-\epsilon,\epsilon)$ so by IVT it should be zero at some point, and the only possibility is that $H_i(0)=H_j(0)$. $\blacksquare$ Claim. $4$ is not interesting. Proof. Assume that $4$ is interesting, then for the permutation $\sigma(4)=3, \sigma(3)=1, \sigma(2)=4$ and $\sigma(1)=2$ there exist polynomials $P_i$ for $1\leq i \leq 4$ such that the three conditions are satisfied. The first condition implies that we can write the polynomials as $P_i(x)=xQ_i(x)+c$ for some real $c$, this makes the other two conditions $Q_1(x)<Q_2(x)<Q_3(x)<Q_4(x)$ for $x\in (-\epsilon, 0)$ and $Q_3(x)<Q_1(x)<Q_4(x)<Q_2(x)$ for $x\in (0, \epsilon)$. Now, applying the lemma on the pairs of polynomials $(Q_3, Q_1), (Q_3, Q_2)$ and $(Q_2, Q_4)$ we have that $Q_1(0)=Q_2(0)=Q_3(0)=Q_4(0)$. Note that we can shift the polynomials such that $Q_1(0)=0$ and without changing the problem. Let $Q_i=xR_i(x)$, now we have that $R_1(x)>R_2(x)>R_3(x)>R_4(x)$ for $x \in (-\epsilon,0)$ and $R_3(x)<R_1(x)<R_4(x)<R_2(x)$ for $x\in (0, \epsilon)$. We again apply the lemma on the polynomials $(R_1,R_2), (R_1,R_4)$ and $(R_4,R_3)$ and get $R_1(0)=R_2(0)=R_3(0)=R_4(0)$. So, from the polynomials $P_i$ we got $R_i$ which satisfy $\deg R_i=\deg P_i-2$ and the same three conditions, so we can assume that one of them is constant and by shifting we can make it $0$, note that the other $3$ can't be constant, but with identical arguments(using the lemma) we can see that $x^n\mid R_i$ for all $n \in \mathbb{N}$, which is impossible. $\blacksquare$ Claim: $2$ is interesting. Proof. Trivial. Claim. $3$ is interesting. Proof. We let $P_i(x)=xQ_i(x)+c$. if $\sigma(i)=i$, we need to find polynomials such that $Q_i$ 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 is clear that linears will work. if $\sigma(1)=1,\sigma(2)=3, \sigma(3)=2$, take $Q_1(x)=0$ and $Q_2,Q_3$ linears going through $0$ and negative leading coefficient. Other permutations can be shown that work, see below for the other cases. $\blacksquare$

As a supplement for @above: Construction for $n=3$: if $\sigma=id$, let $P_1(x)=3x^2$, $P_2(x)=2x^2$,$P_3(x)=x^2$. if $\sigma(1)=1,\sigma(2)=3, \sigma(3)=2$, let $P_1(x)=-x^2, P_2(x)=-2x^3, P_3(x)=-x^3$ and $\epsilon=\frac{1}{2}$. if $\sigma(1)=2,\sigma(2)=1, \sigma(3)=3$, let $P_1(x)=-2x^3, P_2(x)=-x^3, P_3(x)=-x^2$ and $\epsilon=\frac{1}{2}$. For other $3$ permutations $\sigma$, it can be represented as $\sigma(k)=4-\tau(k)$ where $\tau$ is a permutation that has been mentioned above. Let $P_i(x)=-xQ_i(x)$, where $Q_i$s are corresponding polynomials for $tau$. Besides, it's more director to let $P(x)=-xQ(x)+c$ in the proof for $n=4$.

Maybe too easy for Iran TST #3. [Claim] There are 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 there exists an example. Since $P_i (x)-P_1(x)$ also satisfies the inequality. Assume $P_1(x) = 0$. Then, by condition 1, every $P_i (x)$ has constant term zero. Next, $P_3(x)$ is always smaller than $0$ near $0$. Therefore, the minimal term of $P_3(x)$ is some negative number times even power of $x$. For $P_2$ and $P_3$, the minimal term should be some positive number times odd power of $x$. Let $P_2 (x)= a_2 x^{2b_2 -1}+...$, $P_3 (x)=-a_3 x^{2b_3 } +...$, $P_4 (x) = a_4 x^{2b_4 -1}+...$. all $a_i$ are positive real number and $b_i$ are natural number. Since $P_2>P_3>P_4$ for $x<0$, $2b_2 -1 > 2b_3 > 2b_4 -1$. Therefore $b_2 > b_3 \ge b_4$. Then $P_2 <P_4$ for infinitesimal positive real numbers, contradicting the assumption that $P_2 >P_4 > 0 > P_3$. $\blacksquare$ For $n=2, 3$ including zero polynomial the construction is easy.

How about finding all the permutation that such $P_i$ exists? Maybe this corresponds to some graph theoretical facts. As we see above, any case containing $a \rightarrow b \rightarrow d \rightarrow c \rightarrow a$ for $a<b<c<d$ is impossible.