I wrote:
I haven't got a proof, but I suspect 2N-3 crossings is always best.
I still haven't proved anything, but I wrote a brute-force program, and I'm beginning to suspect there's less here than meets the eye! Namely, that for N>1 there are essentially floor((N-2)/2) strategies, depending on how many of a_(n-1), a_(n-3), ..., a_(3 or 4) are greater than 2a_2 - a_1. Then you do 1 and 2 across, 1 back, two largest across, 2 back, that many times, and 1 and largest across, 1 back for the rest, plus a final 1 and 2 across. I've been trying to figure out a way of making this nontrivial, but it starts getting baroque. You have N guys and a (J,K) litter--a litter that holds J guys and needs K other guys to carry it, traveling at the speed of the slowest bearer.... Dan