Clearly, it suffices to show that there exists a permutation of $b_{2n+1}, \cdots, b_0$ such that $b_{2n+1}x^{2n+1} + \cdots + b_1x + b_0$ is strictly increasing. To do this, we will first make use of the following lemma. $\mathbf{Lemma.}$ $x^{2n+1} + x^{2n} + \cdots + x + 1$ is strictly increasing. $\mathbf{Proof.}$ This is clear for positive reals, so we will only show it for negative reals. The derivative of the polynomial is: $$\left (\frac{x^{2k} - 1}{x-1} \right )' = \frac{(x-1)(2kx^{2k-1}) - (x^{2k} - 1)}{(x-1)^2},$$ and so it suffices to show that $2k(x-1)x^{2k-1} > x^{2k} - 1,$ i.e. $2kx^{2k-1} < x^{2k-1} + x^{2k-2} + \cdots + x + 1 \Leftrightarrow x^{2k-1} < \frac{1}{2k-1} * (1 + x + x^2 + \cdots + x^{2k-2})$ (since we're considering negative $x$). However, this is trivial, as $1 + x + x^2 + \cdots + x^{2k-2} > 0$ (by doing casework on whether $x < -1$ or $x \geq -1$, and then pairing terms) and $x^{2k-1} < 0.$ $\blacksquare$ Now, we can simply select the permutation $\{b_i\}$ such that $b_1, b_3, \cdots, b_{2n+1}$ are the $n+1$ largest terms and $b_0, b_2, \cdots, b_{2n}$ are the $n+1$ smallest terms. This finishes because $x^{2i+1}$ is strictly increasing on the negatives whereas $x^{2i}$ is strictly decreasing on the negatives, and because we can write $b_{2n+1}x^{2n+1} + \cdots + b_1x + b_0$ as $c(1+x+x^2 + \cdots + x^{2n+1})$ + (some odd polynomial with all positive coefficients) + (some even polynomial with all negative coefficients), and due to the lemma. $\square$

Choose $b_{2n+1},b_{2n-1},...,b_{1} \ge b_{2n},b_{2n-2},...,b_{2}$. We prove $Q (x)$ is a increasing or $Q'(x)=(2n+1)b_{2n+1}x^{2n}+...+2b_{2}x+b_{1} > 0$ Case 1: $x \ge 0$. $Q'(x)>0$ Case 2: $x <0$ denote $y=-x $. $Q'(x)=(2n+1)b_{2n+1}y^{2n}+...-2b_{2}y+b_{1}=$ $(n+1)b_{2n+1}y^{2n+1}+ \sum_{k=1}^{n}(kb_{2k+1}y^{2k}-2kb_{2k}y^{2k-1}+kb_{2k-1}y^{2k-2})=$ $(n+1)b_{2n+1}y^{2n+1}+\sum_{k=1}^{n}kb_{2k+1}y^{2k-2} \left( (x-\frac{b_{2k}}{b_{2k+1}})^2+\frac{b_{2k+1}b_{2k-1}-b^2_{2k}}{b^2_{2k+1}} \right) >0$ Because $b_{2k+1} \ge b_{2k}, b_{2k-1} \ge b_{2k}$