Suppose I already know the order of i cards, where 3^(k-1)=<i<3^k. I find the number of questions needed to know the position of an (i+1)th card relative to the i cards. If k=1, then it takes only 1 question. Suppose that it takes m questions where k=m. Then for k=m+1, I use one question in this way: I pick two cards so that there are at most (3^m)-1 cards less than the smaller of the chosen cards, between the two chosen cards, and more than the larger chosen card. I can do this since I have at most 3^(m+1)-1 cards. Then I pick the extra card. Now I will know which group that card belongs to, and it will take m further question to determine its exact location relative to the rest of the cards. So, by induction, it will take k more questions to add one more card to our knowledge of relative positions. Then it will take 1+(2*3)*2+(2*3^2)*3+(2*3^3)*4+(2*3^4)*5+123*6=1831 questions to determine the relative order of the 365 cards. Which is what we want. Is this correct?

The same problem with $ 400$ cards and $ 2035$ questions was Argentina IMO TST 2006 problem 2

yes it is always posible.suppose that $f(n)$ be the least number of questions that required for find orders of $n+1$ number such that we know order of $n$ of them,$g(n)$ be the minimum number of such questions that require for finding order of $n$ number,so it is obvious that $g(n+1)\leq g(n)+f(n)$. now prove that $f(n)=f(\left \lceil \frac{n-2}{3} \right \rceil)+1$ and then we find that $g(365)\leq 1828$ and the desired result.

I think my proof is different from the ones above, it's kind of like a triple merge sort from computer science. hopefully I didn't make some silly mistake. Suppose we have $3n$ cards. We divide the cards into $3$ groups and sort each of them independently, so that we obtain one group $a_1<a_2...<a_n$, another $b_1<b_2...<b_n$ and the last group $c_1<c_2...<c_n$. So I ask a question about $a_1,b_1,c_1$, and whichever is the smallest must be the smallest in the entire list, WLOG $a_1$, then I ask a question about $a_2,b_1,c_1$ so whichever is the smallest out of these will be the next smallest in the entire list, and I can keep adding $1$ element in order, and with $3n-2$ questions the entire list will be in order (since for the last $3$ I add them all at once). Now just keep applying the above strategy and you get the recursion $f(3n)=3f(n)+3n-2$ I'm too lazy to count for $365$, but counting for $405$ isn't too hard so I just did that and you need to know that for $5$ cards you can do it in $4$ questions (the strategy is not hard) For $405$ it would take $403 + 3(133) + 9(43) + 27(13) + 81 (4) = 1864$, and so for $365$ we can do it in less than $2000$ questions for sure. Edit: I manned up and did the calculation for $365$, it gives $1655$ questions, now its just a question of whether or not my proof is right. I should also mention the proof by Yossarian is essentially like a binary search but ternary, so I guess both strategies come from computer science.

Pedram-Safaei wrote: yes it is always posible.suppose that $f(n)$ be the least number of questions that required for find orders of $n+1$ number such that we know order of $n$ of them,$g(n)$ be the minimum number of such questions that require for finding order of $n$ number,so it is obvious that $g(n+1)\leq g(n)+f(n)$. now prove that $f(n)=f(\left \lceil \frac{n-2}{3} \right \rceil)+1$ and then we find that $g(365)\leq 1828$ and the desired result. How do you prove that?