ignore this

Sorry, the question seems a little unclear. Are you asking what's the minimum number of pairs of paintings necessary for F.A. to compare such that A.A. can resolve all the other remaining pairs?

Yes, exactly Is the last sentence clear now?

The problem is still not totally clear. Suppose FA marks $A$ better than $B$, $B$ better than $C$, and $C$ better than $D$. Is it allowed that AA can mark $A$ is better than $D$? The problem statement implies he cannot, which seems strange.

You're right, it's not clear by the original Turkish statement too, but we were told he can, as I mentioned in post #2 I translated the original statement directly. If it is necessary, it can be edited properly by admins

The answer is 2014. Represent the paintings and pairs as a graph. That we may need to mark 2014 pairs is obvious. If all pairs of paintings are given, then FA needs to pick enough pairs so that his pairs form a connected graph. Otherwise, AA can't compare $A$ and $B$ where $A$ and $B$ are in different components. Now we show 2014 pairs is sufficient. WLOG, the graph is connected; the disconnected case can be handled by taking components separately (we'll only need $2015 - C$ pairs when there are $C$ components). Have FA perform the following algorithm: 1. Arbitrarily declare some painting $X$ to be the worst, assigning it a score of 1. FA adds it to a set $S$ of paintings he has judged. 2. Take the current highest scored painting $A$ that FA is not yet tired of looking at. 3. If there is no pair $(A,B)$ with $A \in S$ and $B \notin S$, then FA decides he is tired of looking at $A$ and never considers it again. He goes back to step 2. 4. Otherwise, for one of these pairs $(A,B)$, FA decides $B$ is better than $A$ with a score of $|S| + 1$ (best painting so far!). FA then adds $B$ to $S$. 5. FA nods to himself that at this point, (i) the graph formed by paintings in $S$ and FA's selected pairs is a tree, (ii) FA is tired of all the paintings that are not on the unique path from $X$ to $B$, and (iii) any pair $(B,C)$ with $C \in S$ can be resolved by AA. 6. FA goes back to step 2 (and $B$ will be the new $A$). Proof of claim 5(i): Obvious, since every pair FA uses is between one in $S$ and one not in $S$. It's not possible to form cycles, and the graph will always be connected. Proof of claim 5(ii): If $Y$ were a painting not on the path from $X$ to $B$ that FA was not yet tired of, then let $Z$ be the last common vertex in the unique path from $X$ to $B$ and from $X$ to $Y$. FA used $Z$ in an iteration of step 4 despite $Y$ having a higher score and FA not being tired of it, a contradiction of how step 2 and 3 performed. Proof of claim 5(iii): FA can't be tired of $C$ because the edge $(B,C)$ could have been added at the time that he got tired of it. By 5(ii), $C$ must lie on the path between $X$ and $B$. If we take the portion of this path between $C$ and $B$, we get a sequence of pairs that AA can use to resolve $(B,C)$. At the end of the algorithm, taking claim 5(iii) over all points in time guarantees that all pairs on the board can be resolved by AA, and 5(i) means we ended up with a tree of 2015 vertices and 2014 edges. So FA only needs to mark 2014 pairs, as desired.

Answer: $2014$. Example: $(P_1,P_2),(P_1,P_3),..,(P_1,P_{2015}).$ Imagine the paintings as vertices and the pairs which are chosen by participants as edges. Claim $1$: In a connected and undirected graph $G$ we can create a directed spanning tree ,which directs some edges in $G$, such that if there exists an edge between two vertices $A$ and $B$, then there exists a directed path between $A$ and $B$. Proof: Let's do induction on the number of vertices. Let there be $n$ vertices. It's true for $n=1,2$. Assume that it's true for $n=k$. We will prove that it's alos true for $n=k+1$. In $G$ (with $k+1$ vertices), there exists a vertex where the graph stays connected when it's erased (For example we can take a vertex with degree $1$ in the spanning tree of $G$.). Denote this vertex as $A$. We can direct the new graph because of our induction hypothesis. Let the vertices be arranged as $v_1>v_2>...>v_k$. If $A$ is connected with $v_{a_1}>...>v_{a_i}$, then we direct $v_{a_i}\rightarrow A$. So if there exists an edge between $v_{a_j}$ and $A$, then there is a path between $v_j$ and $A$ which is $v_{a_j}\rightarrow v_{a_j+1}\rightarrow v_{a_j+2}\rightarrow...\rightarrow v_{a_i}\rightarrow A$. Our induction is completed. Claim $2$: With an undirected graph $G$, we can create a directed graph $S$ where all the directed edges were undirected edges in $G$ and there exists a directed path between two vertices which had an edge between them in $G$; by directing at most $n-1$ vertices. Proof: Let's do induction on the number of vertices. It's true for $n=1,2$. Assume that it's true for $n=k$. We will prove that it's also true for $n=k+1$. If the graph is connected, then this is true by Claim $1$. If the graph is unconnected, let the components be $C_1,...,C_k$. By Claim $1$, we get that we can direct the edges in $C_i$ with $C_i-1$ directed edges. So we can direct $G$ with $|C_1|+...+|C_k|-k=n-k\leq n-2$ directed edges so the claim is true. Claim $2$ gives that $2014$ directed edges are enough for directing all the vertices as desired.$\blacksquare$