I've had to make 2 assumptions to produce a solution:
1) A tie (1-1 or 2-2) is not a majority, so bottom man gets tossed; and
2) A pirate will vote "no" if he can get the same amount of gold later - he needs a BETTER offer than he can get elsewhere to vote yes.
Given that, pirate #5 makes out big time, suggesting
Pirate 5 - 997 gold pieces
Pirate 4 - 0 gold piece
Pirate 3 - 1 gold pieces
Pirate 2 - 0 gold pieces
Pirate 1 - 2 gold pieces
Working backwards, with 1 pirate, he takes all the money.
So he could have 1000 pieces himself.
With 2 pirates, Pirate #1 votes no to anything, tosses #2, and keeps it all to himself.
With 3 pirates, pirate #3 makes the suggestion. Offering pirate #2 ANYTHING AT ALL will be a win for #2, so the offer is 999 for pirate 3, 1 for pirate #2, and zip for burly-boy #1. #1 votes no, of course, but #2 takes his 1 gold piece and avoids being tossed with everyone else. So if it gets down to 3 pirates, the split is 999/1/0, with a 2-1 vote in favor.
With 4 pirates, pirate #4 ups the ante by 1 for #1 and #2, and cuts #3 out, which gives 997/0/2/1. Pirates #1 and #2 vote for this along with #4, because they know that leaving it to a later round guarantees them less. Pirate #3, of course, votes no. The vote is 3-1 in favor.
Pirate #5 therefore suggests a 997/0/1/0/2 split. #5 of course votes yes, #2 and #4 howl and vote no. #3 looks ahead 1 round, sees himself getting nothing, and accepts the 1 gold piece. #1 looks ahead and sees his take diminish by 1 for each round he waits, so also agrees, making the vote 3-2 in favor.
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This is Mission Control, in Houston.