Can we permute the coefficients of a polynomial so that it has NO real roots?

Yes: put the $n+1$ largest coefficients on the even powers of $x$, and the $n$ smallest coefficients on the odd powers of $x$.

Clearly the polynomial will have no nonnegative roots regardless of the permutation. Changing $x$ to $-x$, it suffices to show: if $\min\{a_{2k}\} \ge \max\{a_{2k+1}\}$, then when $x>0$,$$a_{2n}x^{2n} - a_{2n-1}x^{2n-1} + \cdots + a_2x^2 -a_1x+a_0$$is always positive.

• If $x\ge1$, this follows from $$ (a_{2n}x^{2n} - a_{2n-1}x^{2n-1}) + \cdots + (a_2x^2 -a_1x) +a_0 \ge 0 + \cdots + 0 + a_0 > 0. $$
• If $0<x\le1$, this follows from \begin{multline*} (a_0 - a_1x) + (a_2x^2-a_3x^3) + \cdots + (a_{2n-2}x^{2n-2}-a_{2n-1}x^{2n-1}) + a_{2n}x^{2n} \\ \ge 0 + \cdots + 0 + a_{2n}x^{2n} > 0. \end{multline*}