Problem 1288

1. Let the total amount of money won by the pirates be $x$. This is also equal to the total amount lost by the pirates, since the game is zero-sum.
2. Consider a pirate who has lost a total amount of money $a$. We will show that this pirate can distribute his losses in such a way that he ends up paying exactly $a$.
3. At some stage, let this pirate donate $\frac{99}{100}a$ money to all other pirates. Since there are 99 other pirates, each pirate receives $\frac{99}{100}a \div 99 = \frac{a}{100}$.
4. Now, consider a pirate who has won a total amount of money $b$. We will show that this pirate can receive his winnings in such a way that he ends up receiving exactly $b$.
5. At some stage, let this pirate receive $\frac{b}{100}$ money from all the other pirates. Since there are 99 other pirates, each pirate pays $\frac{b}{100} \div 99 = \frac{b}{100}$.
6. At the end of the process, the first pirate (who lost $a$) has donated a total of:
$$\frac{99}{100}a + \frac{1}{100}a - \frac{1}{100}x + \frac{1}{100}x = a$$
This shows that the pirate who lost $a$ has paid exactly $a$.
7. Similarly, the second pirate (who won $b$) has gained a total of:
$$\frac{99}{100}b - \frac{1}{100}x + \frac{1}{100}b + \frac{1}{100}x = b$$
This shows that the pirate who won $b$ has received exactly $b$.

Therefore, it is possible for each winner to receive exactly what he has won and for each loser to pay exactly what he has lost.

$\blacksquare$