9.4. A confectionery factory produces $N$ types of candies. For New Year, the factory gave each of 1000 school students a gift containing candies of several types (the compositions of the gifts could be different). Each student noticed that for any 11 types of candies, they received a candy of at least one of these types. However, it turned out that for any two types, there is a student who received a candy of exactly one of these two types. Find the largest possible value of $N$.
Answer. $N=5501$.

Solution. Let $A_{1}, A_{2}, \ldots, A_{N}$ be the sets of students who did not receive candies of the 1st, 2nd, ..., $N$-th types, respectively. According to the problem, all these sets are distinct; moreover, each student is contained in no more than ten of them. Therefore, the total number of elements in these sets does not exceed $1000 \cdot 10 = 10000$.

Suppose there are exactly $k$ one-element sets among our sets. Then the number of sets containing more than one student does not exceed $\frac{10000-k}{2}$, so the total number of all non-empty sets does not exceed $\frac{10000-k}{2} + k = \frac{10000+k}{2} \leqslant \frac{11000}{2} = 5500$. Taking into account that one of the sets can be empty, we get that $N \leqslant 5501$.

It remains to show how the described situation could arise when $N=5501$. Make the set $A_{5501}$ empty; the distinct sets $A_{4501}, \ldots, A_{5500}$ will each contain one student. It remains to choose distinct two-element sets $A_{1}, \ldots, A_{4500}$ such that each student is in exactly 9 of them. For this, for example, we can divide the students into 100 groups of 10 people and take all pairs of children in the same group (there will be exactly $100 \cdot \frac{10 \cdot 9}{2} = 4500$ such pairs).

Remark. The construction can be completed in different ways. For example, we can arrange the students in a circle and form each two-element set either from a pair of opposite students (there will be 500 such pairs) or from a pair of students with no more than three other students between them.

Comment. Only the correct answer without justification (or with incorrect justification) - 0 points.

Only proof that $N \leqslant 5501$ - 4 points.

Only an example with $N=5501$ types - 3 points.

If the participant did not take into account that one type can be present in all gifts (i.e., the corresponding set $A_{i}$ can be empty) and therefore obtained the answer 5500 - 1 point is deducted.