Suppose the cdf of your distribution is F, whose inverse I will call G to avoid cumbersome notation. Then your quantity is the limit of 1/(n-1) [G(1/n) + G(2/n) + ... + G((n-1)/n)], which is one Riemann sum on the way to the integral of G. An easy change of variable shows that this integral equals the mean (or doesn't exist if the mean doesn't). So that resolves question 1.
Cheers! That's just what I needed.
In a previous post, I looked at some of the properties of using the median rather than the mean.
Inspired by Househalter's comment, it seems we might be able to take a compromise between median and mean. It seems to me that simply taking the mean of the lower quartile, median, and upper quartile would also have the nice features I described, and would likely be closer to the mean.
Furthermore, there's no reason to stop there. We can take the mean of the n-1 n-quantiles.
Two questions:
Note the unlike the median approach, for large enough n, this maximiser will pay Pascal's mugger.