TITLE: Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups?
QUESTION [53 upvotes]: For a finite group G, let |G| denote the order of G and write $D(G) = \sum_{N \triangleleft G} |N|$, the sum of the orders of the normal subgroups.  I would like to call G "perfect" if D(G) = 2|G|, since then the cyclic group of order n is perfect if and only if the number n is perfect.  But the term "perfect group" is taken, so I'll call such a group immaculate.
My question is:

Does there exist an immaculate group of odd order?

Since the cyclic immaculate groups correspond one-to-one with the perfect numbers, a "no" answer would immediately prove the famous conjecture that there are no odd perfect numbers.  However, perhaps someone can easily see that there is a non-cyclic immaculate group of odd order, proving that the answer is "yes".
Here's what I know.  There are no abelian immaculate groups except for the cyclic ones.  (Edit: more generally, if $D(G) \leq 2|G|$ then every abelian quotient of $G$ is cyclic.  Proof: not hard, and given here.)  However, there do exist nonabelian immaculate groups, e.g. $S_3 \times C_5$ (of order 30).  Derek Holt has computed all the immaculate groups of order less than or equal to 500.  Their orders are
$$
6, 12, 28, 30, 56, 360, 364, 380, 496
$$
(Integer Sequence A086792).  Of these, only 6, 28 and 496 are perfect numbers; the rest correspond to nonabelian immaculate groups.  Some nonabelian immaculate groups of larger order are also known, e.g. $A_5 \times C_{15128}$, $A_6 \times C_{366776}$, and, for each even perfect number n, a certain group of order 2n.  But these, too, all have even order.
Edit: Steve D points out that p-groups can never be immaculate.  This also appears as Example 2.3 here; it follows immediately from Lagrange's Theorem.  I should have mentioned this, as it rules out an easy route to a "yes" answer.

REPLY [14 votes]: Almost a decade late to the party, but here is another example:
$$(C_7 \rtimes C_{3^2}) \times (C_{19^2} \rtimes C_5)  \times C_{11^2 \cdot 197} = \text{SmallGroup}(63, 1) \times \text{SmallGroup}(1805, 2) \times C_{23837}.$$
where $\text{SmallGroup}(a,b)$ denotes the $b$th group of order $a$ according to the database of GAP. Then:
$$\frac{D(C_7 \rtimes C_{3^2})}{|C_7 \rtimes C_{3^2}|}\frac{D(C_{19^2}\rtimes C_5)}{|C_{19^2}\rtimes C_5|}\frac{D(C_{11^2})}{|C_{11^2}|}\frac{D(C_{197})}{|C_{197}|} = 
\frac{95}{63}\cdot \frac{2167}{1805}\cdot \frac{133}{121}\cdot \frac{198}{197} $$
$$=\frac{5 \cdot 19}{3^2\cdot 7} \cdot\frac{11 \cdot 197}{5\cdot19^2}\cdot \frac{7 \cdot 19}{11^2}\cdot \frac{2 \cdot3^2\cdot 11}{197}=2.$$
This group is of order $2710624455$ and is thus about 100 times smaller than François Brunault's example.
I found this group for a project for one of my courses. Immaculate groups seem to be very rare in general, examples of order are even harder to find. For further results, see  http://www.math.ru.nl/~bosma/Students/JorisNieuwveld/A_note_on_Leinster_groups.pdf