$100$ people have $100$ one-dollar bills. Some give bills to others until all have different amounts. What's the least number of people to give money? 
In a group of $100$ people, each of them has a wallet with $100$ one-dollar bills. Some of them gave one or more dollars to one or more of the others and eventually everyone ended up with a different amount $>0$. What is the least number of people who gave some of their money?

I have started by trying to make $10,000$ as a sum of different integers. One way is $4+6+7+\cdots+141$, which means that $95$ people have given money to others but I don’t think this is the correct approach.
Also tried taking one person's amount and sharing it to others, so for example, $4+5+\cdots+14 = 99$ but again this will take for ever and we won't know if it's the optimal.
Can anybody help me out? Thank you!
 A: For everyone to have a different amount of money, everyone (except perhaps one person) has to either give or receive some money. Without loss of generality we may assume that nobody is both receiving and giving money; if person $A$ receives money from person $B$ and gives money to person $C$, then we get the same result if person $B$ gives money directly to person $C$.
To minimize the number of people giving money, we must maximize the number of people not giving money. Every pair of people not giving must receive a different amount of money. To minimize the amount of money given, the first receiver receives $\$0$, the next receiver receives $\$1$, etcetera, until everyone not giving has received money (possibly $\$0$). The total amount received is then
$$0+1+2+\ldots+(n-1),$$
where $n$ is the number of people not giving money. To minimize the number of people giving this amount of money, the first giver gives $\$99$, the next giver gives $\$98$, etcetera, until there is sufficient money to give to everyone.
So now you want to find the maximal positive integer $n$ such that
$$0+1+2+3+\ldots+(n-1)\leq 99+98+97+\ldots+n.$$
The number of people giving money is then $100-n$.
