Bump~~~~

Probably I've made some mistake, but I've found this solution... Take the usual graph interpretation of the problem and induct in the number $e$ of edges. The base case $e=1$ is trivial. For the inductive step, let $G$ be a graph with $e$ edges and let $v$ be a vertex of $G$, with degree at least 1, and let $\Gamma (v)$ be the friends of $v$. If $G'$ is the graph obtained by removing $v$ from $G$ and $G''$ is the graph obtained by removing $\Gamma (v)$ from $G'$ , then we have 2 cases: 1. If $G''$ is empty, simply put $v$ and $u\in \Gamma(v)$ in the golden list, in this order. 2. If $G''$ have at least 1 vertex, we partition $G''$ into connected components. Let $A$ be the set of connected components with at least 1 edge, $B$ be the set of connected components with 1 vertex, each of then connected with some vertex of $\Gamma (v)$, and let $C$ be the remaining vertices. Since the vertices in $C$ aren't connected with any vertex in $G$, we can simply ignore then, so we'll suppose that the vertices of $G''$ are only in $A$ or $B$. We have 2 more cases. 2.1. If $B$ is empty, then by hypothesis we have a golden list $L=v_1,...,v_{2k}$ to $G''$ with a even number of vertices (because $A$ isn't empty). To make a golden list to $G$ with the same property, simply put $v$, $v_1,...,v_{2k}$ and $u$, in this order ($u\in \Gamma(v)$). $v$ writes all $\Gamma(v)$ in the blackboard, each $v_i$ writes at least one vertex in $G''$ by hypothesis, and $u$ writes $v$. Therefore, in this case we're done. 2.2. Se $B$ isn't empty, let $w\in B$ and note that, in $G$, $w$ is connected with $u$ iff $u\in \Gamma(v)$. Now, by hypothesis, there's a golden list $L=v_1,...,v_{2k}$ to $G'$. Take $w\in B$ such that $w$ is the first one to appear on the blackboard if we follow the golden list $L$. Since $w$ was written on the blackboard, some $u\in \Gamma(v)$, friend of $w$, is in the golden list. But $u$ also write $v$ in his turn, so all the vertices of $G$ are in the blackboard if we follow $L$. Therefore, $L$ is a golden list to $G$, and we're done in this case, if there's no isolated vertex $t \in \Gamma(v)$ in $G'$, i.e., if $deg(t) = 1$ in $G$. For this case, we'll find another way to construct the golden list. 3. Let $\Gamma(v) = X \cup Y$, where $X$ is the set of vertices with degree 1 ($t\in X$) and $Y$ the set of vertices with degree at least 2. Clearly, every $w \in B$ is connected to some vertex $u \in Y$. Let $u_1, ...,u_p$ vertices in $Y$ such that, if we put then in the golden list, in this order, then every vertex in $B$ will appear on the blackboard (at least one of then appear in each step). Let $L=v_1,...,v_{2k}$ be the golden list to $A$. 3.1. If $p$ is odd, let the golden list to $G$ be $v,v_1,...,v_{2k},u_1,...,u_p$. It's easy to see that it works. 3.2. If $p$ is even, let the golden list to $G$ be $t, v, v_1, ...,v_{2k}, u_1, ..., u_p$. It's easy to see that it works. Therefore the problem is solved by induction.

We will show friends of $x$ with set $S(x)$. Take a random student and call it $x$. Write it in the first place of golden list. Take a student from the group which is written on the blackboard by $x$ and write this student in the second place. Since this is taken from $S(x)$ we can write it down. If every student who has at least one friend is written on the blackboard then we are done. Assume contradiction and take a student $y$ and write it in the third place of golden list. In case we can write $y$ in the golden list we will continue to do the same thing. This is our algorithm: take a student $v$ which has not been written on the blackboard, write it down if you can and for the next place of golden list take a student which has just been written on blackboard by $v$. This will work unless we can't find a student $v$. Let's write until we can and because of pairing students in algorithm we will have students $s_1, s_2, ..., s_{2k}$ on golden list and we are stuck. Take student $u$ and we know that it is not written on blackboard, because of algorithm again. Thus we know we can take one friend of $u$ -let's call $w$- such that it can be written on golden list after $s_{2k}$ and any of $u$'s friends is not among $s_i$s. If we can create the golden list $u, s_1, s_2, ..., s_{2k}, w$; we will be done. Assume contradiction. We know $w$ can write $u$, so we do not need to worry about it. Take the smallest $i \in {1, 2, ..., 2k}$ such that $s_i$ cannot write a new friend after operation putting $u$. If $i=2m$, then that means $s_{2m}$ cannot write $s_{2m-1}$ on the blackboard, which means $u \in S(s_{2m-1})$. Contradiction. On the other hand, if $i=2m-1$, then due to the fact that $s_{2m} \in S(s_{2m-1})$, we can say $s_{2m} \in S(u)$ but at first we said that $u$ was not written by any of $s_i$s. Contradiction. Finally we created a new list and we are not stuck. Thus we can keep going until everyone who has at least one friend is written on the blackboard which gives us a golden list.

Take the obvious graph interpretation. We proceed by induction on number of vertices. WLOG no vertex has degree 0. Pick a random vertex $u$ and consider the induced subgraph of $G\backslash (\{u\} \cup N(u))$ If this induced graph has no vertices of degree 0, we can start with the list for the graph, add u (which is okay because vertices in N(u) will be added) and a vertex in N(u) (which adds u) If the induced subgraph has a vertex of degree 0, call v. Note $N(v)\subseteq N(u)$ because $uv$ is not an edge or $v\in N(u)$ and after removing $N(u)$, $v$ has no neighbors. If $N(v)=N(u)$ I can consider $G'=G\backslash u$. A golden list for $G'$ is a golden list for $G$ because at least one neighbour of v is in the list, which adds u. Otherwise, if $N(v)\subseteq N(u)$ then consider a golden list with first vertex being u and another golden list starting with $u,v$. After these moves the set of vertices on the blackboard are the same, but these lists' lengths must have different parities.

@CANBANKAN you said If this induced graph has no vertices of degree 0, we can start with the list for the graph, add u (which is okay because vertices in N(u) will be added) and a vertex in N(u) (which adds u) ofcourse you can write a vertice in N(u), but...can we really add u? if every vertice in N(u) have a friend in the golden list of G\u,N(u),then everyone in N(u) was written on the board before you write u,then u cannot be in the list,how can you solve this problem?

@above You are right, but this has an easy fix since you can just take $u$ first, then the golden list for $G\backslash (\{u\} \cup N(u))$ and then the vertex from $N(u)$. See that this list satisfies.