Solution. Suppose one of the students has received 1001 tickets, each with the number $\frac{1000}{1001}$. Since the sum of the tickets is 1000, this student must be able to distribute the tickets over the 100 boxes. By the pigeonhole principle, there must be at least one box with 11 tickets. The sum of these 11 tickets is $11 \cdot \frac{1000}{1001}=11\left(1-\frac{1}{1001}\right)=11-\frac{11}{1001}=11-\frac{1}{91}$. We will now show that this is the smallest possible value, $c=11-\frac{1}{91}$.

For any student, we first choose the distributions in which the maximum of the sums of the tickets per box is as low as possible, and from these distributions, we then choose a distribution in which the fewest number of boxes achieve this maximum. Let $d_{1} \leq d_{2} \leq \ldots \leq d_{100}$ be the total value of the 100 boxes in this distribution, ordered from low to high (where the last $k$ are equal to the maximum). Since the sum of all tickets is at most 1000, we note that

$$
99 d_{1}+d_{100} \leq d_{1}+d_{2}+\ldots+d_{100} \leq 1000 .
$$

On the other hand, moving a positive ticket from box (with sum) $d_{100}$ to box (with sum) $d_{1}$ should not yield a better distribution by assumption: thus, no distribution with a lower maximum or with fewer than $k$ boxes of this maximum value. This means that the new value of $d_{1}$ must be at least $d_{100}$. If $d_{100} \leq 10$ we are done, because $1010$. Since each ticket is at most 1, this means that box $d_{100}$ must contain at least 11 positive tickets. This, in turn, implies that there is a ticket in this box with a positive value of at most $\frac{d_{100}}{11}$. If we move this ticket to box $d_{1}$, then it must hold that

$$
d_{1}+\frac{d_{100}}{11} \geq \text { "new value of box } d_{1} " \geq d_{100} \text {. }
$$

This can be rewritten as $11 d_{1} \geq 10 d_{100}$. If we combine this with the first equation, we find

$$
91 d_{100}=90 d_{100}+d_{100} \leq 99 d_{1}+d_{100} \leq 1000
$$

For every student, the lowest maximum of the sums of the tickets per box is thus $d_{100} \leq \frac{1000}{91}=\frac{1001}{91}-\frac{1}{91}=11-\frac{1}{91}$.
Task 2. Every student in the Netherlands receives a finite number of cards. Each card has a real number in the interval $[0,1]$ written on it. (The numbers on different cards do not have to be different.) Find the smallest real number $c>0$ for which the following holds, regardless of the numbers on the cards that everyone has received.

Every student whose sum of the numbers on the cards is at most 1000 can distribute the cards over 100 boxes so that the sum of the numbers in each box is at most $c$.