A question on the sum of element orders of a finite group Let G be a nontrivial finite group. Is it true that the sum of the orders of all elements of G is not divisible by the order of G? 
 A: It is false in general, for instance there's a group of order $3\cdot 5\cdot 7=105$ with sum of orders equal to $1785=3\cdot 5\cdot 7\cdot 17$. (In Magma, it is the first of the two groups of order 105 in the "small groups" database).
However it is true for all groups of even order, because the sum of orders of elements is always odd (this is shown by partitioning $G$ according to the equivalence relation $x\sim y$ if $x$ and $y$ generate the same cyclic subgroup, and using the fact that, for a positive integer $n\geq 1$, $n\varphi(n)$ is odd only if $n=1$.)
A: Edit: I misunderstood the question, I'll try to fix here. I don't have the complete answer but I'll try to give a partial answer:
let $G$ be a group of order $|G|$ and for each $d \mid |G|$ let $n_d$ indicate the number of elements of order $d$ in $G$; then if $|G|$ is even $|G| \nmid \sum_{d \mid |G|}n_d d$.
Indeed we have that if $d$ is a odd divisor of $|G|$ (not equal to $1$) either $n_d=0$ or exists a odd prime numeber $p$ such that $p-1 \mid n_d$ and so $n_d$ is even, on the other hand if $d$ is even clearly $n_d d$ is also even and so $\sum_{1 \ne d \mid |G|} n_d d$ must be even.
Thus $\sum_{d \mid |G|}n_d d$ is odd and so $|G| \nmid \sum_{d \mid |G|} n_d d$, because by hypothesis $|G|$ is even.
