I'm assuming each person starts with 0 apples.
A less creative solution for your above example with the same number of steps is:
B gives A 250 (A at 250, B at -250)
C gives B 120 (B at -130, C at -120)
D gives C 420 (C at 300, D at -420)
E gives D 320 (D at -100, E at -320)
F gives E 340 (E at 20, F at -340)
Proceeding inductively, it can be shown that given n people with 0 apples can be rearranged to have the above requirement which totals 0 (n-1) trades is a solution. To prove that it is the least amount of trades, suppose that any Ending Requirement can be yielded with (n-2) trades instead. Let n=2; it's obvious that end result A 1 B -1 cannot happen with 2-2=0 trades, so we have a contradiction. Hence the smallest number of trades possible to get the required result is (n-1)
EDIT: the quantity of each trade will be (projected value - current value) of the receiver after each trade.