After considering this problem hard, I finally found the following. Let $a_i$ be the card written $i$. The dealer write the following one hundred messages; $a_1$ beats $a_{100}$, $a_3$ beats $a_1$, $a_{100}$ beats $a_3$, $a_3$ beats $a_2$, $a_4$ beats $a_3$, $\ldots$, $a_{99}$ beats $a_{98}$. Then, we show that only $a_i=i$ for any $i=1,2,\ldots,100$ is possible. Since the first three messages are "trilemma", two of $a_1, a_{100}, a_3$ contains 1 and 100. But $a_3$ is beaten by two others and beats two others, it cannot be 1 or 100. Thus, $a_1=1, a_{100}=100$. From other 97 messages, $a_2, a_3, \ldots, a_{99}$ are sorted order and none of them are 1 or 100, we must have $a_2=2, a_3=3, \ldots, a_{99}=99$.

There is no way where $n=3$. Actually, 3 information is sufficient to find exact numbers where $n=4$ : $a_{2}$ beats $a_{1}$, $a_{3}$, and $a_{4}$ beats $a_{3}$. If $a_{2}$ is $4$, $a_{4}=1$ beats someone except $4$ which is contradiction. So $a_{2}$ must be $3$. It leads to conclusion that $a_{1}=1, a_{2}=3, a_{3}=2, a_{4}=4$. Unfortunately I found the ineffective way where $n=5$, i can't help but trying to using inductive ways.(if $n=5$, $a_{1}<a_{2}<a_{3}<a_{4}>a_{1}$, $a_{3}>a_{5}$) Let's suppose induction holds where $4<k<n$, It says that we could find sub-graph containing $k$ edges which have only one way to label $k$ vertices. in case of $k=n$, we erase one number except $1,n$. By inductive hypothesis, there is sub-graph meeting the condition. After you erase a number, let's see direct of edges between erasing number and the remaining in every case. If you get sub-graph, you can find a unique di-graph by comparing $k$ with some number of di-graph. Actually, you always get unique di-graph by erasing $2$ because that number is the smallest except $1$.

We write $i>j$ iff $i$ beats $j$ CLAIM. The only solution to the system \begin{align} a_{100}&<a_{1}\\ a_{3}&<a_{100}\\ a_1&<a_3\\ a_2<a_3&<a_4<...<a_{99} \end{align}is $a_i=i$ for every $1\leq i\leq 100$. Proof. Firstly if $2\leq a_1,a_3,a_{100}\leq {99}$ then from $a_1>a_{100}$ and $a_{100}>a_3$ we have $a_3>a_1$. Therefore at least one of $a_1,a_3,a_{100}$ is in the set $\{1,100\}$. It is easy to see that the only possibility is that one of $(a_{100},a_1),(a_3,a_{100}),(a_1,a_3)$ is equal to $(100,1)$. Therefore, we have $2\leq a_2,a_4,...,a_{99}\leq 99$. Now suppose on the contrary that $a_3=1$ then $a_3>a_2$ where $a_2\neq 100$ which is impossible, similarly $a_3\neq 100$. Hence we must have $(a_{100},a_1)=(100,1)$. Now from $2\leq a_2,a_3,a_4,...,a_{99}\leq 99$ and $(4)$ we easily obtain $a_i=i$ for every $1\leq i\leq 100$$\blacksquare$ Since the dealre knows the order of the card, he can offer the message which consists of the above system and we are done.