4. On his birthday, Piglet baked a large cake weighing 10 kg and invited 100 guests. Among them was Winnie-the-Pooh, who is fond of sweets. The birthday boy announced the rule for dividing the cake: the first guest cuts a piece of the cake that is $1 \%$, the second guest cuts a piece of the cake that is $2 \%$ of the remaining part, the third guest cuts a piece of the cake that is $3 \%$ of the remaining part, and so on. What place in line should Winnie-the-Pooh take to get the largest piece of cake?
Solution. The first guests in the queue receive increasingly larger pieces of the pie because the remaining part of the pie is large at the initial stages of division. However, since the remaining part of the pie decreases, there will come a point when guests start receiving smaller pieces of the pie. At which guest will this happen?

Let's compare the pieces of the pie received by the $(n-1)$-th and $n$-th guests. Suppose the mass of the remaining part of the pie when it was the $(n-1)$-th guest's turn is $m$ kg. Then the $(n-1)$-th guest received $\frac{n-1}{100} \cdot m$ kg, and the remaining part of the pie became $m - \frac{n-1}{100} \cdot m$ kg, which simplifies to $\frac{101-n}{100} \cdot m$ kg. Therefore, the $n$-th guest received $\frac{101-n}{100} \cdot m \cdot \frac{n}{100} = \frac{n(101-n)}{10000} \cdot m$ kg. The difference between the pieces received by the $n$-th and $(n-1)$-th guests is $\frac{n(101-n)}{10000} \cdot m - \frac{n-1}{100} \cdot m = \frac{-n^2 + n + 100}{10000} \cdot m$. This difference is positive if $n^2 - n - 100 < 0$. Solving the quadratic inequality, we find that the largest natural solution is $n=10$. Therefore, the 10th guest in line will receive the largest piece of the pie, so Winnie-the-Pooh should take the 10th place in the queue.

Answer: 10.