At the final state, let $P=\{p_{1},p_{2},...,p_{m}\}$ be the set of students who have at least one card, $Q=\{p_{m+1},...,p_{n}\}$ be not. In this state, any student in $P$ has posted at least $n-2$ times. $X$ is the he number of posting $P$ to $Q$, $X=(n-m)m-L$ ($L$ is the number of students who never posted to one student in $Q$). Let the $Y$ be the total sum of the posting $Q$ to $P$, the $C_{0}(p_{i})$ be the number of cards whose $p_{i}$ have initially, it holds $Y\ge \sum_{i=m+1}^{n}C_{0}(p_{i})+X$. Now that $C_{0}(p_{i})\ge1$, $Y\ge(n-m)(m+1)-L$. It leads to conclusion that $L\ge n-m$ because $(n-m)m\ge Y$. Now suppose that there is a Tree $(t_{1},...,t_{x})$ such that $t_{i}$ doesn't give card to $t_{i+1}$. If $t_{x}\in Q$, and the others in $P$, $t_{i}$ should have $t_{i+1}$ cards only. So there should be someone who has $t_{1}$'s cards, and he has given cards $n-1$ times. Let the set of students who gives $n-1$ times be $F$. If $|F|>0$, students in $F$ should have their own cards and one of the $t_{1}$. However, they have given $n-1$ times so that the number of cards in $F$ can't be over than initial state. Contradiction. Let the $k$ students $p_{1},...,p_{k}$ satisfied second problem's condition be the "Bad cycle". As discussed above, there is no "Tree" in final state. This implies that students is classified some Bad cycle, and $F$ as mentioned. any student in Bad cycle gave and brought $n-2$ times exactly.(because there is no Tree) So, students in each part have to has cards from same part. in Bad cycle especially, $C(p_{i})=C_{0}(p_{i+1})$($C(p_{i})$ is the number of cards of $p_{i}$ in final state). So we done.

This is inspired from roughlife's idea. Let $S$ be the set of all students. Let the `final state' refer to the case that no one can post the greeting cards. At the final state, let $P$ be the set of students with at least one card and $Q$ be the complement of $P$ relative to $S$. Let $G$ be the directed graph with vertex set $S$. For each pair of distinct students $s, t \in S$, either $s$ have posted exactly one card to $t$ or $s$ have not posted any card to $t$. Let $G$ has the edge $s \to t$ if and only if $s$ have not posted any card to $t$. For every $p \in S$, let $I(p)$ and $F(p)$ be the initial and final number of cards which $p$ possesses, respectively. Let $\Delta (p) = F(p)-I(p)$. Let $\hbox{deg}^-(p)$ be the indegree (the number of incoming arrows of $p$) and $\hbox{deg}^+(p)$ be the outdegree (the number of outgoing arrows of $p$). Observing that $p$ has posted $n-1-\hbox{deg}^+(p)$ cards and received $n-1-\hbox{deg}^-(p)$ cards, we have \[ \Delta(p) = \hbox{deg}^+(p) - \hbox{deg}^-(p) \] We can observe that each vertex in $P$ has at most one arrow with tail ends on it(outgoing arrow); if it has more than two outgoing arrow, then the corresponding student can choose an envelops to post his/her card which contradicts to the condition of final state. Moreover, the following holds. ($\ast$) For every $p \in P$, if $p$ has an outgoing arrow from $p$ to $q$, the only card $p$ has is those with sign of $q$. Hence for every $p \in P$, there is a unique walk from $p$ and exactly one of the following happens (a) The walk enters a (non-trivial) loop in $P$. (b) The walk ends at a final point in $p' \in P$ (including the case that $p'=p$, i.e. there is no outgoing arrow from $p$.). (c) The walk enters $Q$. Therefore $P$ is partitioned into three sets $A$, $B$, $C$ which consists those $p \in P$ corresponds to the case (a), (b), (c) above, respectively. Note that \[ \sum_{p \in B} F(p) \ge \sum_{p \in B} I(p) \]since for every $p \in B$, card with $p$s sign cannot exist in $A$ or $C$ or $Q$. Because of $\ast$, all the cards finally in $A$ were initially in $A$. Similarly, the only cards finally in $C$ were initially in $Q$ or $C$. Now it will be explained that the union of all cards in $B$ is preserved. As stated, $\sum_{p \in B} \Delta p = \sum_{p \in B} F(p) - I(p) \ge 0$. On the otherhand, \[ \sum_{p \in B} \Delta p = \sum_{p \in B} \hbox{deg}^+(p) - \hbox{deg}^-(p) \]Note that on the right hand side, for all internal arrows within $B$, the indegree and outdegree cancels out. There is no arrow from $B$ to $A$, $C$, or $Q$. There is no arrow from $A$ or $C$ to $B$. Therefore, \[ \sum_{p \in B} \Delta p = -\textrm{the number of arrows from $Q$ to $B$} \le 0 \]Hence the number of union of all cards in $B$ is preserved. As observed, all cards initially in $B$ will stay at the final state in $B$. Therefore the set of all cards initially in $B$ is preserved; any other cards cannot enter, no cards initially in $B$ can escape. Situation in $A$ is simpler. It will be explained that in $A$ there is no internal initial point. This means that $A$ is partitioned into disjoint loops.(Since there cannot be a fork, there cannot be any crossing) To see this, suppose that $A$ has an initial point $p$. Then cards with $p$s sign should be found somewhere else. But it cannot be found in $A$ or $B$ or $C$ or $Q$. Hence there is not initial point. Therefore each $p \in A$ has the same in and out degree, $1$. Hence $F(p) = I(p)$ for all $p \in A$. Also note that cards with signs of $p \in A$ cannot found anywhere else. Hence the multiset of cards in $A$ were preserved. Finally consider $C$. Suppose that $C \neq \emptyset$. Let $q \in Q$ be a vertex which is directly reachable from some $p \in C$. As before, by the backward chasing the walk from $q$, we reach a point $p \in C$ which does not have any incoming arrow from $C$. Then the card with sign of $p$ cannot found anywhere else which leads to a contradiction. Therefore $C = \emptyset$ which implies $\sum_{p \in P} \Delta(p) = \sum_{p \in C} \Delta(p) = 0$. Then \[ \left| Q \right| \le \sum_{q \in Q} I(q) = -\sum_{q \in Q} \Delta (q) = -\sum_{s \in S} \Delta (s) =0 \]Thus we have $Q = \emptyset$. Now we have $S$ is a disjoint union of $A$ and $B$. It remains to show that for any loop $\left(p_1 , p_2 , \ldots , p_k\right)$ in $A$, $I(p_j) = I(p_{j+1})$, $j=1,2, \ldots , k$, $p_{k+1} = p_1$. As mentioned, $\Delta(p_j) = 0$ for all $j$. Finally $I(p_{j+1})= F(p_{j+1}) = I(p_{j})$ completes the proof.