Take $a_1 < 0$, and the remaining $a_i = 0$. Then $E_n = a_1(n-1) < 0$ for $n$ even, so the proposition is false for even $n$. Suppose $n \ge 7$ and odd. Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4= b$, and $a_5 = a_6 = ... = a_n = c$. Then $E_n = (a - b)^3 (a - c)^{n-4} < 0$. So the proposition is false for odd $n \ge 7$. Assume $a_1 \ge a_2 \ge a_3$. Then in $E_3$ the sum of the first two terms is non-negative, because $a_1 - a_3 \ge a_2 - a3$. The last term is also non-negative. Hence $E_3 \ge 0$, and the proposition is true for $n = 3$. It remains to prove $S_5$. Suppose $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$. Then the sum of the first two terms in $E_5$ is $(a_1 - a_2){(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)} \ge 0$. The third term is non-negative (the first two factors are non-positive and the last two non-negative). The sum of the last two terms is: $(a_4 - a_5){(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)} \ge 0$. Hence $E_5 \ge 0$.

Here is an intellectually honest solution to this problem. Let $p(x) = (x - a_1)...(x - a_n)$ then $E_n = p'(a_1) + ... + p'(a_n) $. If $p$ has repeated roots $a_i=a_j$ then $p'(a_i) = p'(a_j) = 0$. Note also that if $a$ is root of $p$ with multiplicity $m$ then it will stay on the same side of $x$ if $m$ is even but will cross it if $m$ is odd. Further if $a$ is a singular root then $p'(a) < 0$ if the polynomial passes from the positive side of $x$ to the negative side and $p'(a) > 0$ when it passes from negative to positive. This now informs thinking about the different cases. (For clarity let's assume $a_1 \leq a_2 \leq ... \leq a_n$). If $p$ is even then the polynomial starts in the upper plane and can pass to the lower plane next to a singular root (where $p'$ is negative), and then go back to the upper plane via the remaining (odd number of) repeated roots. Then if $p$ is odd with degree larger than $7$, then it starts in the lower plane can pass to the upper plane via a $3$ times repeated root, pass back to the lower plane via a singular root and then back to the positive plane via an odd number of repeated roots. The remaining two cases can then be verified via size considerations.