A non-detailed sketch of solution. Replace $k=2024$ which is even. Take the graph interpretation. Let's prove that $N\geq \lceil log_2k\rceil$. Denote the set $S_i$ such that for the sets $A_i$ and $B_i$ which are created after $i.$ partition, WLOG $|S_{i-1}\cap A_i|\geq |S_{i-1}\cap B_i|$. Define $S_i=S_{i-1}\cap A_i$. (Take the larger set after every partition.) $|S_i|\geq \lceil \frac{k}{2^i}\rceil$ Hence, $N\geq \lceil log_2k \rceil$. Let's prove that we can organise $N$ balanced tournaments where $N\geq \lceil log_2k \rceil$. Divide vertices into two sets $X$ and $Y$ with equal number of members. In a partition if we divide $X$ into $a,b$ then we also divide $Y$ into $a,b$ and we group $(a,b)$ where we get equally sized sets.$\blacksquare$ Example: $\{1,2,3,4,5,6\}\rightarrow (\{1,2,3\}),(\{4,5,6\})\rightarrow (\{1,2\},\{6\}),(\{3\},\{4,5\})\rightarrow (\{1\},\{4\},\{6\}),(\{2\},\{3\},\{5\})$ Remark by erkosfobiladol: Let $k=2^n+2m$ such that $0< m<2^{n-1}$.Choose $2^{n-1}+m$ different binary codes consisting of $n$ digits. Write $1$ after their last digits. We have $2^{n-1}+m$ different binary codes with $n+1$ digits. Take their conjugates, then we have $2^n+2m$ binary codes. In $i.$ move, construct two sets such that the codes with their $i.$ digit $1$ are in the first set, and the codes with their $i.$ digit $0$ are in the second set. These partitions satisfy the conditions given in the problem.(If $m=0,$ then consider all binary codes with $n$ digits and apply the last procedure.) Example of Remark: $11,10,01\rightarrow 111,101,011\rightarrow 111,101,011,000,010,100$ We have $(111,101,100),(011,000,010), \ (111,011,010),(101,000,100) \ (111,101,011),(000,010,100)$.

anantmudgal09 wrote: A league consists of $2024$ players. A round involves splitting the players into two different teams and having every member of one team play with every member of the other team. A round is called balanced if both teams have an equal number of players. A tournament consists of several rounds at the end of which any two players have played each other. The committee organised a tournament last year which consisted of $N$ rounds. Prove that the committee can organise a tournament this year with $N$ balanced rounds. Proposed by Anant Mudgal and Navilarekallu Tejaswi We may rephrase the problem as covering edges of the complete graph $K_{2024}$ by complete bipartite subgraphs of type $K_{x, 2024-x}$ as a tournament and doing so by only using subgraphs $K_{1012, 1012}$ as a balanced tournament. Label the vertices of $K_{2024}$ by $1, 2, \dots, 2024$. Suppose we have a tournament with rounds $G_1, \dots, G_N$, each $G_i$ being the disjoint union of sets $A_i$ and $A_i^c$ inside $\{1, 2, \dots, 2024\}$. Define $$\Delta = \sum_{i=1}^{2024} \left||A_i|-|A_i^c|\right |.$$We show that while $\Delta>0$, we can find another tournament with $N$ rounds while decreasing $\Delta$; once $\Delta = 0$, we get a balanced tournament. Indeed, pick a $G_i$ and assume without loss of generality $|A_i| < |A_i^c|$; we will try to transfer a $u \in A_i^c$ to $A_i$ to form a new round. This is always possible provided every edge from $u$ to members of $A_i$ occurs in another one of the rounds. If we successfully transfer $u$, $\Delta$ decreases and we're done, so we may instead assume there is a vertex $v_u \in A_i$ such that the edge $uv_u$ is not covered in any of the rounds. This must be true for all vertices $u \in A_i^c$, else, we can transfer some vertex $u$, and $\Delta$ decreases as desired. Since $|A_i^c|>|A_i|,$ by pigeonhole principle, there exists $u, w \in A_i^c$, and $v \in A_i$ such that $v=v_u=v_w$. Thus neither of the edges $vu, vw$ are in any of the rounds. So in each of the other rounds, $v, u$ and $v, w$ are on the same side of a partition, hence $u, w$ are always on the same side in each of the other rounds. Thus, $uw$ edge is never covered in any of the rounds including $A_i$! This is a contradiction. Thus the algorithm terminates with $\Delta=0$ as desired.

where are the rest of the problems?

Why cant us non pdcers take it

Consider the graph with players as vertices. Two vertices are connected by an edge if they have not yet played together. The graph starts out at a complete clique and must end as a null graph. At any point the graph consists of a collection of disjoint cliques. After a round a cliques is broken into two parts determined by which players play on which team. We show that $11\leq N$ and there is a balanced construction for $N=11$. Bound: By induction after $n$ rounds their must be a clique of size at least $2024\cdot 2^{-n}$, thus $11\leq n$. Construction: We prove that in one round there exists a balanced tournament that reduces the maximal clique size from $m$ to $\lceil m/2 \rceil$. Notice there must be an even number of odd cliques. If a clique has an even number of players split them equally amongst the two teams. If a cliques has an odd number of players then split them almost equally amongst the teams with an equal number of odd cliques giving their extra player to each team.