Answer. The maximum $N$ is 25, and there exist two different types of families of 25 subsets that satisfy the condition of the problem.

Solution. Consider an arbitrary family of $N$ distinct non-empty subsets such that each element of the set $X$ is contained in exactly two subsets of this family. If there are $x$ one-element subsets, then the total sum $\mathrm{S}$ of the cardinalities of the subsets in this family is at least $x + 2(N - x) = 2N - x$. On the other hand, each element of the set $X$ is contained in exactly two subsets of this family, so $\mathrm{S} = 34$, which means $\mathrm{N}$ does not exceed $(34 + x) / 2$. Finally, the number of one-element subsets $x$ does not exceed the number of elements in $X$, which is 17, so $N$ does not exceed 25. Note that for $N = 25$, the number of one-element subsets in the family can be 16 or 17.

1) If $x = 17$, then the remaining 8 "large" subsets must be two- or more-element subsets. If there are $y$ two-element subsets among them, then the total number of elements in the 8 large subsets is 17 and is at least $2y + 3(8 - y) = 24 - y$, so $y$ is at least 7. If it is greater than 7, then the three- and more-element subsets in total would contain one element, which is impossible. Therefore, the only possibility is 7 pairwise disjoint two-element subsets and one three-element subset. Together with the 17 one-element subsets, they form the first example of the family of subsets sought in the problem. An example of a family of this type: 17 one-element subsets $\{1\}, \{2\}, \ldots, \{17\}$, 7 two-element subsets $\{1, 2\}, \{3, 4\}, \ldots, \{13, 14\}$, and one three-element subset $\{15, 16, 17\}$.
2) If $x = 16$, then the remaining 9 subsets must be two- or more-element subsets. If there are $y$ two-element subsets among them, then the total number of elements in the 9 large subsets is 18 and is at least $2y + 3(9 - y) = 27 - y$, so $y$ is at least 9. Therefore, all large subsets are two-element subsets. In total, they contain 18 elements, so two of them intersect at one element (which does not lie in any one-element subset), while the rest are pairwise disjoint. Together with the 16 one-element subsets, they form the second, different from the first, example of the family of subsets sought in the problem. An example of a family of this type: 16 one-element subsets $\{1\}, \{2\}, \ldots, \{16\}$, 9 two-element subsets $\{1, 2\}$, $\{3, 4\}, \ldots, \{13, 14\}, \{15, 17\}, \{16, 17\}$.

In the course of the reasoning, we proved that there are no other families different from the ones found.

Grading criteria. For each found example of one of the two types of families: 1 point each. Proof of the maximality of $N = 25$: 3 points. Proof that there are exactly 2 types of maximal families: 2 points.
11.4. In a set $X$ of 17 elements, a family of $N$ distinct non-empty subsets is selected such that each element of the set $X$ is contained in exactly two subsets from this family. What is the maximum value of $N$? Find the number of all possible different types of such families for the maximum $N$. Two families of subsets have different types if one cannot be obtained from the other by permuting the elements of $X$.