what is the cycle length of the maximum normalized cycle in the directed complete graph? Consider the complete, directed graph on $n$ vertices. Let the edge lengths $\{X_{ij}: 1 \leq i, j \leq n\}$ be i.i.d standard normal, with the constraint $X_{ij} = -X_{ji}$. The value of a normalized cycle is the sum of the edges involved, divided by the cycle length. We want to know: 
For any fixed $k$ and large $n$, (of particular interest are $k = 3$ and $k = n/2$), what is the probability that the maximum normalized cycle is of length $k$? 
Some thoughts: 
Note that $3 \leq k \leq n$. There are $\binom{n}{k}(k-1)!$ directed cycles of length $k$ (except for $k =2$, in which we have $\binom{n}{k}$ such cycles), and each normalized cycle of length $k$ is Gaussian with variance $\frac{1}{\sqrt{k}}$. For small $k$ and large $n$, the number of directed cycles of length $k$ is approximately $n^k$. For $k = n/2$, this number is approximately $\sqrt{2}(\frac{2n}{e})^k$. Therefore, cycles of small length has the advantage of having larger variance, cycles of longer length has the advantage that there are many more of them. 
To see that the dependency between cycles really matter, suppose that all cycles are independent. Since max of $m$ i.i.d Gaussian is $\sqrt{2\pi \log m}$, for small $k$, we have
$$E(\max \mbox{cycle of length k}) \approx \frac{2\sqrt{2\pi k \log(n)}}{\sqrt{k}} = 2\sqrt{2\pi \log(n)}$$. 
For $k = n/2$, we have
$$E(\max \mbox{cycle of length n/2}) \approx \frac{2\sqrt{2\pi k (\log(n) + \log(2/e)}}{\sqrt{k}} = 2\sqrt{2\pi (\log(n) + \log(2/e))}$$. 
But the max of $m$ i.i.d Gaussians with variance $\frac{1}{k}$ has variance $\approx \frac{1}{k}$ (Borell's inequality), therefore the difference of $\sqrt{\log (2/e)}$ will not get picked up. 
Another naive approach: consider an easy union bound to get an upper bound on $E(\max \mbox{cycle of length k}) $. Let $Z$ denotes the standard normal. Then
$$
P(\exists \mbox{ a $k$-cycle } > m) \leq \binom{n}{k}(k-1)!\cdot P(Z > \sqrt{k} m) \leq \exp(k\log n - k\log C - \frac{1}{2}\log(k) - \log(m) - \frac{km^2}{2})
$$
where $C$ is some fixed constant. Solve for $m$ so that the RHS is $1$, we see that $m \approx \sqrt{2\log n}$, so this is an uninformative bound. 
So one needs to take into account the dependency between cycles. But I'm quite stuck on what to do here. A quick literature search didn't return anything useful. Any ideas will be appreciated. 
Thanks!
 A: This was too long for a comment.  (I ignored edge directionality - not sure it matters)
Taking the maximum of i.i.d. variables reduces the variance. As $k$ is increased, you take the maximum of a larger number of r.v.s., each already with a smaller variance, so the overall variance is further reduced. Say $y_N$ is the maximum of $m$ i.i.d. Gaussians each $N(0,\sigma^2)$, you have $E[y_N] \sim  \sigma \sqrt{2\log m}$ and $V[y_N] \sim \sigma/\sqrt{2\log m}$. Using a (very) crude approximation $\binom n k = n^k$ you can approximate the mean and variance of the maximal cycle of length $k$: the mean is $E_k \sim \sqrt{2 \log n}$ so independent of $k$ (similar to what you've got), but the variance $V_k \sim \frac{1}{k \sqrt{2 \log n}}$ does depend on $k$. Therefore, at least under the i.i.d approximation, larger cycles give a smaller variance thus smaller $k$'s are more likely to give the maximum ($k=2$ the most likely). 
A: I'm hoping someone will take these thoughts and run with them, so that eventually the question gets answered and stops bumping to the top.
Let's set n=4 and look at 3 cycles vs 4 cycles. Arrange labeling so that 4 of the normalized values are for the 3-cycles are a,b,c,and d, and so that the following relations hold: a+b+c+d=0, and the values for 3 of the 4-cycles are 3/4 times one of (a+b), (a+c), or (-b-c).  If a is the largest value, then b and c have to be smaller than a/3 and their sum bigger than -4a/3 in order for the 3 cycles to win the prize for maximal normalized value.  One might be able to work out the probabilities for this case, and then make a similar comparison between pairs of k cycles and (2k-2l) cycles for judicious choices of k, l, and n.
My intuition on this is poor, but it suggests to me that 4 cycles have a slight edge on 3-cycles for n=4 and even for larger n.
It may be possible to build up a set of inclusion-exclusion type relations for n+1 based on the relations for n.
Gerhard "Someone Take The Baton Now" Paseman, 2012.08.07
