Problem 711

Solution. Let $100 S$ be the total amount of money in the world (regardless of the currency), and the total number of people be $100 N$ (where $N$ can be non-integer, which is not important). Then at least $90 S$ of the money is owned by $10 N$ people (let's call this group oligarchs). The remaining $90 N$ people (let's call them commoners) have less than $10 S$ in total. Let's take the poorest half of the commoners. These $45 N$ people have no more than half of the money of all commoners, which is no more than $5 S$. Therefore, if we take all people except these $45 N$, they will have at least $95 S$ in total. There are $55 N$ such people, which is $55 \%$. We have proven that $55 \%$ of the richest people own $95 \%$ or more of all the money.

Let's provide an example showing that less than $55 \%$ may not be possible. Suppose there are 100 people, 10 of whom have 81 million each, and 90 have 1 million each. Then out of the total capital of 900 million, $90 \%$ (810 million) belongs to the first ten people, and the condition of the problem is satisfied. $95 \%$ of the capital is 855 million. To accumulate this amount, it is necessary to have the 55 richest people: $10 \cdot 81 + 45 \cdot 1$.

Answer: $55 \%$.