Answer: $n=7$. We will present an algorithm for how to indicate $2^{n-1}$ sets on $2n$ cards, one of which is suitable. (In our case, this is $2^{6}=64<80$ sets.) Imagine that identical cards are connected by (invisible to us for now) red edges. We will arbitrarily pair the cards with blue edges. The red-blue graph is a collection of cycles, and any set of the form "cards every other in each cycle" will suit us. Any such set has the obvious property: on each blue edge, exactly one vertex is selected. Therefore, by checking all $2^{n}$ sets with this property, we will certainly find the suitable one. Moreover, all such sets are divided into pairs of opposites (and in each pair, either both sets are suitable or both are not), so it is enough to check $2^{n-1}$ sets.

Now suppose that for 16 cards, 80 sets of 8 cards each are indicated. We will prove that it is possible to write numbers on the cards so that none of these sets are suitable.

For each set, mark all $8^{2}$ pairs of cards where one card is in the set and the other is not; in total, $80 \cdot 8^{2}$ pairs will be marked. There are $16 \cdot 15 / 2$ pairs in total, so some pair will be marked no more than $80 \cdot 8^{2} /(16 \cdot 15 / 2)=(8 / 15) \cdot 80$ times. Therefore, this pair is "split" by no more than 42 sets. We will write the number 1 on this pair of cards. All sets that contain both of these cards or neither of them are already rejected. We will now consider only the remaining $a \leqslant 42$ sets that potentially could be suitable.

Each of them contains exactly 7 of the remaining 14 cards. Similarly to the previous step, some pair of them will be "split" by no more than $[(7 / 13) \cdot a] \leqslant[7 / 13] \cdot 42]=22$ sets. We will write the number 2 on this pair and thereby reject all sets except no more than 22.

We continue in the same manner, eventually leaving 4 free cards and one set of two cards. We will write the number 7 on the cards of this set, and the number 8 on the two remaining cards; this set will also be rejected.
7. There are $2 n$ cards, each with a number from 1 to $n$ (each number appears on exactly two cards). The cards are lying on the table face down. A set of $n$ cards is called good if each number appears exactly once. Baron Munchausen claims that he can point out 80 sets of $n$ cards, at least one of which is guaranteed to be good. For what largest $n$ can Baron's words be true?