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For the sake of contradiction, assume that this is not possible.

So, there is not $k$ boxes with $\frac{100}{k}$ or more balls because taking these $k$ buckets keeping exactly $\lceil\frac{100}{k}\rceil$ balls satisfies the condition.

•  So, there is no box with 100 or more balls.
•  there is not 2 boxes with 50 or more balls.
•  there is not 3 boxes with 34 or more balls.
•  there is not 4 boxes with 25 or more balls.
• there is not 5 boxes with 20 or more balls.
• There is not 100 or more boxes with balls.

So, the highest amount of total balls is $99+49+33+24+19\cdot(99-4)=2010<2023$

Contradiction.

So, this is always possible.