Essentially, consider this operation: Pirate A gives $x$ gold to all other pirates, and then Pirate B takes $x$ gold from all other pirates. After these operations, the net effect is that Pirate A has given $(n+1)x$ gold to Pirate B ($n$ is the number of pirates), leaving all other pirates' gold amounts unchanged. By changing $x$ you can easily make $(n+1)x$ any positive real you want, so now you can make one pirate give an amount of gold to another. The remaining part is clear. To prove it simply, pick out a pirate M whose win or loss is smallest in magnitude, pick any other pirate N who won if M lost or lost if M won. Transfer gold between M and N so that M has the right amount of gold. Use induction on the pirates minus M.

Let the pirates who have won have won a total of money $x$. This is also equal to the total amount lost by the pirates who have lost. Consider a pirate who has lost a total of money $a$. At some stage, let this pirate donate $\dfrac{99}{100}a$ money to all other pirates. Now, consider a pirate who has won a total of money $b$. At some stage, let this pirate get $\dfrac{b}{100}$ money from all the other pirates. At the end of the party, the first pirate has donated a total of $\dfrac{99}{100}a + \dfrac{1}{100} x + \dfrac{1}{100}a - \dfrac{1}{100}x = a$, and the second pirate has gained a total of $\dfrac{99}{100}b -\dfrac{1}{100} x + \dfrac{1}{100}b + \dfrac{1}{100} x = b,$ so the job is possible.