Suppose one of the pupils has been given $1001$ cards, each containing the number $\frac{1000}{1001}$. Since the sum of the cards is $1000$, this pupil should be able to distribute the cards among the $100$ boxes. Because of the pigeonhole principle, there is at least one box with $11$ cards. The sum of these $11$ cards is $11 \cdot \frac{1000}{1001} = 11(1 - \frac{1}{1001}) = 11 - \frac{11}{1001} = 11 - \frac{1}{91}$. We are now going to show that this is the smallest possible value, i.e. $c = 11 - \frac{1}{91}$.

For a random pupil, we first consider those distributions for which the maximum of the sums of the cards per box is as small as possible. From these distributions we then pick a distribution for which the number of boxes having their sum equal to this maximum, is as small as possible. Let $d_1 \le d_2 \le \dots \le d_{100}$ be the sums corresponding to the $100$ boxes in this distribution, ordered from low to high (with the last $k$ of them equal to the maximum). Since the sum of all cards is at most $1000$, we have that
$$99d_1 + d_{100} \le d_1 + d_2 + \dots + d_{100} \le 1000.$$
On the other hand, moving a positive card from the box (with sum) $d_{100}$ to the box (with sum) $d_1$ cannot create a better distribution per assumption: i.e. the new distribution does not have a smaller maximum, or less than $k$ boxes equal to this maximum value. This means that the new value of $d_1$ is at least equal to $d_{100}$. If $d_{100} \le 10$ we are immediately done, because $10 < 11 - \frac{1}{91}$. So we may assume that $d_{100} > 10$. Since each card is at most $1$, this implies that box $d_{100}$ contains at least $11$ positive cards. This in turn implies that there is a card in this box with positive value at most $\frac{d_{100}}{11}$. Therefore, if we move this card to box $d_1$, then it must hold that
$$d_1 + \frac{d_{100}}{11} \ge \text{``new value of box } d_1\text{''} \ge d_{100}.$$
We can rewrite this as $11d_1 \ge 10d_{100}$. Combining this with the first equation, we find
$$91d_{100} = 90d_{100} + d_{100} \le 99d_1 + d_{100} \le 1000.$$
So, for each pupil, the smallest maximum of the sums of the cards per box is $d_{100} \le \frac{1000}{91} = \frac{1001}{91} - \frac{1}{91} = 11 - \frac{1}{91}$. $\square$