Source: IMO-2017-P5
Category: Existence and Construction

Problem:
Fix $N \ge 1$. A collection of $N(N+1)$ soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove $N(N-1)$ players from this row to obtain a new row of $2N$ players in which the following $N$ conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, \dots, no one stands between the two shortest players. Prove that this is possible.

Solution:
We begin by partitioning the $N(N+1)$ players into $N$ groups based on their heights. Order the players by height and define the groups as follows:
Let $G_1$ consist of the $N+1$ shortest players, 
$G_2$ consist of the next $N+1$ shortest players, 
and so on, up to $G_N$, which consists of the $N+1$ tallest players. 

If we select exactly two players from each group $G_k$, then in the final row of $2N$ players, the two players from $G_N$ will be the 1st and 2nd tallest, the two from $G_{N-1}$ will be the 3rd and 4th tallest, and so on, down to the two from $G_1$ being the two shortest. Thus, to satisfy the problem's conditions, it suffices to select two players from each group $G_k$ such that for each $k$, the two selected players from $G_k$ end up adjacent in the final row of $2N$ players.

We prove that this is always possible by proving a more general claim by induction on $M$:
**Claim:** *For any integer $M \ge 1$, given a row of players and $M$ disjoint groups of players such that each group has at least $M+1$ players present in the row, we can select two players from each group such that, after removing all unselected players, the two selected players from each group are adjacent in the remaining row.*

**Proof of Claim:**
We proceed by induction on $M$.

*Base case:* For $M=1$, we have $1$ group with at least $2$ players. We simply select the first two players from this group in the row. Since all other players are removed, these two players will trivially be adjacent.

*Inductive step:* Assume the claim holds for some $M-1 \ge 1$. We are given $M$ groups, each with at least $M+1$ players. 
Scan the row of players from left to right, and stop as soon as we encounter two players belonging to the same group. Let this group be $G_k$, and let the two players be $A$ and $B$, where $B$ is the player we just scanned (so $B$ is to the right of $A$). 

By the definition of our stopping condition, the scanned prefix of the row contains exactly two players from $G_k$, and at most one player from each of the other $M-1$ groups. We select $A$ and $B$ as the two players for group $G_k$.

Next, we delete the entire scanned prefix of the row (up to and including $B$), and we also delete all remaining players of $G_k$ from the rest of the row. For each of the remaining $M-1$ groups, we have deleted at most $1$ player (since at most one player from each of these groups appeared in the scanned prefix). Therefore, each of the remaining $M-1$ groups still has at least $(M+1) - 1 = M = (M-1) + 1$ players present in the remaining row.

By the induction hypothesis, we can select two players from each of these $M-1$ groups from the remaining row such that they are adjacent in the final selection. Because $A$ and $B$ are the only players selected from the deleted prefix, they will appear at the very left of the final row of $2M$ selected players. Thus, no other selected player can stand between $A$ and $B$, meaning $A$ and $B$ are adjacent. The pairs selected from the remaining groups are also adjacent by the induction hypothesis. This completes the inductive step.

**Conclusion:**
Applying the claim for $M = N$ to our initial partition of $N$ groups (each of size $N+1$), we can successfully choose two players from each group such that they are adjacent in the final row. As established, this selection satisfies all the conditions of the problem.


Grading guidelines:
(Partial) 1. Made the key reduction by partitioning the players into height groups (e.g. the k-th group consists of the (N+1) players with ranks (k-1)(N+1)+1 through k(N+1)), and observed that it suffices to choose two players from each group so that each chosen pair is adjacent in the remaining row. 2. Stated/proved the inductive auxiliary claim (or an equivalent Hall-type/greedy statement): for M disjoint groups, each appearing at least M+1 times in the row, one can choose two from each group so that within the chosen subsequence the two from each group are consecutive. 3. In the induction/greedy step, correctly identified the “scan left to right until the first repeated group” idea and proved that in the scanned prefix there is at most one element from each other group, hence deleting this prefix reduces the required lower bound for each remaining group by at most 1. (Almost) 1. Gave the full inductive construction selecting a first pair by scanning to the first repeated group, then deleting the scanned prefix and all remaining members of that group, and applying induction to the remaining M−1 groups, but did not fully justify why the selected first pair ends up adjacent in the final row (e.g. did not argue that all other selected players come from the suffix). 2. Correctly completed the auxiliary claim, and correctly applied it to the N groups of size N+1 to finish, but left a small gap in the counting after deletions (e.g. did not justify that each remaining group still has at least (M−1)+1 elements in the reduced row). 3. Solution is almost complete, but made minor mistakes which are not negligible.