The general principals that you need are your basic 'ohms law' which says that the voltage drop across any circuit element is (including the wires) is simply the current times the resistance of that element.
Each of the conductors in this circuit add resistance and cause voltage drop.
The only problem with putting the skinny wire first is that it has to carry _all_ of the current for _all_ of the lights, which means that you will have your highest resistance section where you have your highest current, meaning high voltage drop...but this is also the shortest run.
I am guessing at 24A of current (6000W, 277V, something for power factor).
#6Al has a resistance of 0.808 ohms per 1000 feet. You have 400feet (200 out, 200 back) and so a resistance of 0.3232 ohms. V=IR, so the voltage drop is 7.75V for this first section.
The second section carries 16A. #1Al has a resistance of 0.253 ohms/1000 feet. You have 800 feet and a resistance of 0.2024 ohms with a voltage drop of 3.24V.
The final section carries 8A, 600 feet of wire, 0.1518 ohms, voltage drop of 1.21V.
Total voltage drop of 12.2V or about 4.5%. Kind of on the high side but possibly okay, depending on the way the ballast regulates versus reduced voltage.
I've made the simplifying assumption that the voltage drops are small, and that the lamps draw the same current. This is actually not true, since the current consumption will change with voltage.
The bulk of your voltage drop is in the first section. You might consider doing that stretch in copper; using #6Cu the voltage drop is reduced to 4.8V
-Jon
