https://artofproblemsolving.com/community/c6h355784p1932924

Sorry, forgoten the case that the sets can have only 2 permuations. Thanks @below

@above, 144,225,351,432,513

Easy problem. We use first and second conditions, we get digits are $1,2,3,4,5,6,7$.So, maximum number of this form numbers are less than 7. $$711$$$$621,612$$$$531,513,522$$$$441,414,432,423$$$$351,315,342,324,333$$$$261,216,252,225,243,234$$$$171,117,162,126,153,135,144$$Sum of digits are 9. Case 1:$ n=7 $. So,we must choose exactly one number from each group.This means we must use 711. But if we use 711, then we don't use 621 and 612. Because, we have third and fourth conditions. Case 2: $ n=6 $. Subcase 1: We use 711. We don't use 621,612,531,513,441,414,351,315,261,216,171,117.So, From third group we must choose 522, then we mustn't use 432 or 423.This means $n \le 4$. Subcase 2: We don't use 711. This means we use 621 or 612. If we choose 621, then we must choose 513. And we must choose 432.This means in fifth we don't choose 351,315,342,324,333.But this is not true. Case 612 is similar with case 621. Case 3: $ n=5 $. In this case we have construction. 513,432,351,225,144.(rmtf1111's construction)

The solution I wrote during the exam. Make 3 sets $U$, $T$, $H$, first with the unitary digit of the numbers, second with the tens digit of the numbers, and third with the hundreds digit of the numbers. Sum of all elements in these 3 sets is obviously $9n$, so the sum of elements in the at least one of the sets is less than or equal to $3n$. Let this set be $S$. Also, we know that there are exactly $n$ positive and different elements in the set $S$, so the sum of the elements in the set $S$ is bigger than equal to $1+2+\cdots +n = \frac{n(n+1)}{2}$. Therefore, $n(n+1)\leq 6n$ or $n \leq 5$. Equality case uses only the digits $1,2,3,4,5$ and we can easily find construction with that (look above for that).

I also kinda had the same boundi solution using the sum of the digits of all the numbers as the post above when I tried it, but I'm just wondering, why was this placed as P2 instead of P1?

Sum of all number is 9n but sum is at least n(n+1)/2 which for n>5 is impossible.For n=5 contruction is 144,225,351,432,513.

We find that the answer can be 5. It is the most because if there are more than 5, the increase of sum of digits is 9. But then we at least add 6*3=18, 18>9, contradiction.

It is clear that $9$ and $8$ cannot be used since $8+1+1>9$. If $7$ is used, then one of the numbers must be a permutation of the digits $711$. The only remaining digit combinations that sum to $9$ not containing a $1$ are $(5,2,2)$, $(4,3,2)$, and $(3,3,3)$. If $(3,3,3)$ is used, then no permutation of $(4,3,2)$ can be used. We can only use one permutation of $(5,2,2)$, so together with a number starting with $1$, this can only achieve $1+1+1+1=4$ numbers. If some permutation of $(5,2,2)$ is used, then only one permutation of $(4,3,2)$ can be used, so this also limits the total to $4$ numbers. Finally, if $3$ permutations of $(4,3,2)$ are used, then we cannot use either permutations of $(1,2,6)$, $(1,3,5)$, or $(1,4,4)$, so again this limits the total to at most $4$. Now, if for the contradiction using the digits $1$ through $6$ three times each yields $6$ numbers, then since $6$ can only be used in combination with $2$ and $1$, all three $2$s and $1$s are used up, so $3$ can only be combined with itself. This is bad because $4$ is now unable to be used. Finally, the combination $531$, $432$, $351$, $225$, $144$ works, so the answer is $n=5$.