Whats now this " 1st india-iran friendly compeititon"

Both countries' IMO teams compete in a friendly maths olympiad before the IMO, started this year

reminds me of the type of Discrete* Hardy Littlewood Maximal Function (w.r.t ARO 2000, ChTST 2021 ,USA TSTST 2015P1) For more details check out this:-https://dgrozev.wordpress.com/category/analysis/hardy-littlewood-maximal-function/

Can someone please post a solution

@above I think for the problem, we should first 'greedily' sort the indices {1,2,3,....,n} according to the values of a i^ in descending order and after constructing the set , we can invoke the discrete Hardy Littlewood maximal function in the context of partitioning a sequence of real numbers! (The exact formulation of the problem may not match the standard form of discrete HL maximal function,but it shares similarities in spirit)

bump....

bump....

Hopefully I did not make a mistake... Let $m$ be the minimum of the RHS such that $$m(2k+1-2j)\leq \sum_{i=j}^{n}{a_i}$$for all $1\leq j\leq k$. The case $k=1$ proves itself. The case $n=k$ is also trivial as the sets are just the elements and $$m(2n+1-2j)\leq \sum_{i=j}^{n}{a_i}\leq (n+1-j)a_j\Rightarrow m\leq a_j$$ If it happens that $m\leq a_1$ then we can make a set with just $a_1$ and then one realizes that the rest of the inequalities are the same as the case $ (n-1,k-1)\mapsto (n,k)$ and $j+1 \mapsto j$ so we can induct down on $n$ until we have that all $a_i$ are smaller than $m$. Assign $a_1,a_2,\dots,a_k$ to $B_1,B_2,\dots,B_k$, respectively. We then preform the greedy algorithm, assigning the remaining elements until each set has a sum of at least $m$. The amount of 'waste' in this process is at most $a_{k+1}+a_{k+2}+\dots +a_{\min(2k,n)}$. So it suffices to show that $$\sum_{i=k+1}^{n}a_i-(a_{k+1}+a_{k+2}+\dots +a_{\min(2k,n)})\geq \sum_{i=1}^{k}(m-a_i)$$However we know that $m(2k-1)\leq \sum_{i=1}^{n}{a_i}$ so it suffices to show that $$m(k-1)\geq a_{k+1}+a_{k+2}+\dots +a_{\min(2k,n)}$$If this holds then we are done so assume it does not. If $n<2k$ then we have a contradiction by assumption. Notice that we have that $a_{2k-1}>m/2$ and so we can still come up with a partition if $a_{k+1}\geq 1-a_{2k}$ by pairing elements which clearly must hold.

Amazing problem. Let $a_1+a_2+\dots+a_n=S$. First, we show that if all $a_i<\frac{S}{k}$, then we can group them into $k$ groups so that each sums to $\ge \frac{S}{2k-1}$. In this case, we will be done, simply with the $j=1$ term on the RHS.

Now, suppose there is an element $\ge \frac{S}{k}$. Then, $a_1\ge \frac{S}{k}$. We now put $a_1$ by itself in the partition. Since $a_1\ge \frac{S}{k}$, this means that it is definitely not the smallest amongst $\sum\limits_{i \in B_j} a_i$. Hence, we can effectively remove it from the LHS of the inequality. It suffices to show the exact same inequality with $n$ replaced by $n-1$ and $k$ replaced by $k-1$, and so we are done by induction on $k$.

See also this blogpost.