Problem 5. 99 children are standing in a circle, each initially holding a ball. Every minute, each child throws their ball to one of their two neighbors. If a child is simultaneously thrown two balls, one of them is lost irretrievably. What is the minimum number of minutes after which the children can be left with only one ball?
Solution. We will prove that after at least 98 minutes in the game, only one ball can remain (the ball is thrown at the end, not the beginning, of the corresponding minute). We number the children and the balls from 1 to 99 in the clockwise direction.

We will show how only one ball can remain after 98 minutes. On the first, third, and every subsequent odd move, the first ball is passed from the first child to the second, and on every even move - from the second child to the first. (Other balls that are passed to these children are lost.) Other balls with odd numbers are passed in the opposite direction of the clock on each move. Balls with even numbers are passed in the clockwise direction. Thus, on the first move, ball 3 is lost at the second child, on the second move, ball 98 is lost at the first child, and so on. Balls with odd numbers greater than 1 end up at the second child on odd moves and are lost; the last ball with number 99 is lost on the 97th move. Similarly, balls with even numbers end up at the first child on even moves, simultaneously with the first ball. The last of the even-numbered balls is the second ball, which is lost on the 98th move.

We will show that after $n$ minutes, where $n<98$, the balls are always more than one. For convenience, we will imagine that the balls do not get lost but are glued together (the numbers are preserved). Let's assume that the balls end up with the first child at the end.

Let $n$ be odd. All the balls must end up with the first child after $n$ minutes, i.e., after an odd number of minutes. If the first ball does not go around all the children, it will return to the first child only after an even number of moves. Therefore, the first ball must go around the entire circle, which requires at least 99 minutes.

Let $n$ be even. If the second ball does not make a full circle minus one step, it will end up with the first child only after an odd number of minutes. Therefore, the second ball must go around the circle minus one step, which requires at least 98 minutes.

Therefore, when $n<98$, the balls in the circle are always more than one.

Comment. We received incomplete solutions indicating how to leave only one ball after 98 minutes, but without proof of the minimality of this period.