To determine whether there exist positive reals $a_0, a_1, \ldots, a_{19}$ such that the polynomial $P(x) = x^{20} + a_{19}x^{19} + \ldots + a_1x + a_0$ does not have any real roots, yet all polynomials formed from swapping any two coefficients $a_i, a_j$ have at least one real root, we proceed as follows:

Consider the polynomial $P_\sigma(x) = x^{20} + a_{\sigma(19)}x^{19} + a_{\sigma(18)}x^{18} + \cdots + a_{\sigma(0)}$, for all permutations $\sigma$ of the numbers 0 to 19.

We construct the coefficients $a_i$ in a specific manner. Let $a_i = 10000 + i\epsilon$ for $i = 0, 1, \ldots, 19$ and some small $\epsilon > 0$. This ensures that $a_0 < a_1 < \cdots < a_{19}$.

When $t = 0$, we substitute $x = -100$. Since $\frac{|a_{19} \cdot 100^{19}|}{20} > |100^{20}|, |a_{18} \cdot 100^{18}|, |a_{17} \cdot 100^{17}|, \ldots, |a_0|$, we have $P(-100) < 0$.

As $t \rightarrow \infty$, $a_{18} \rightarrow \infty$. When $a_{18} > -\min_{x < 0} \left( x^2 + a_{19}x + \frac{a_{17}}{x} + \cdots + \frac{a_0}{x^{18}} \right)$, $P(x) \geq 0$ for all $x < 0$. This minimum exists because as $x \rightarrow 0$, $\frac{a_0}{x^{18}}$ dominates and the sum tends to positive infinity, so it is positive for some $x > x_0$. Meanwhile, as $x \rightarrow -\infty$, $x^2$ dominates, and the sum is positive for some $x < x_1$. The middle interval is closed and bounded, so it achieves its minimum which is finite.

Meanwhile, $P(x) > 0$ for all $x \geq 0$.

Fix $t$ as the minimum value such that $P(x) \geq 0$ for all $x$. By continuity, there is a root $y$ of $P(x)$, which is clearly negative. If $-1 \leq y < 0$, then $a_{19}y^{19} + a_{18}y^{18} > a_{18}(y^{18} + y^{19}) \geq 0$. Grouping the rest similarly in pairs, and using $y^{20} > 0$, $P(y) > 0$, a contradiction.

Hence $y < -1$, and $y^{19} < y^{17} < \cdots < y^1 < y^0 < y^2 < \cdots < y^{18}$. Since $a_{19} < a_{17} < \cdots < a_1 < a_0 < a_2 < \cdots < a_{18}$, by the rearrangement inequality, $0 = P(y) > P_\sigma(y)$ for $\sigma \neq \text{Id}$.

Adding a small $\delta$ to $t$, $P(x) > 0$ for all $x$, while $P_\sigma(x)$ ($\sigma \neq \text{Id}$) takes both positive and negative values. Therefore, such positive reals $a_0, a_1, \ldots, a_{19}$ do exist.

The answer is: \boxed{\text{Yes}}.