I think it is a pretty nice problem though I didn't have a solution for it. My first idea towards it is induction,if we use induction hypothesis we can deduce it to $ \sum x_i \ge mn$ similar for $ \sum y _ i$but I can't continue with it. My second idea is pigeon principle. That is if we can find two of $ x_1,x_1 + x_2,...x_1 + x_2 + ... + x_m,y_1 + x_1...y_1 + x_1,y_1 + x_1 + ...x_m,y_1 + y_2 + x_1,...y_1 + y_2 + x_1 + ... + x_m,...y_1 + ... + y_n + x_1 + ... + x_m$ are equal then we are done.So I want to use some ineqs prove it.But I failed Can you continue with it,or present your ideas or solutions

I finally found a solution(I wish it is correct)! But I want to see some other approaches or ideas before posting my solution I will post my solution this weekend

My solution

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hxy09 wrote: That is if we can find two of $ x_1,x_1 + x_2,...x_1 + x_2 + ... + x_m,y_1 + x_1...y_1 + x_1,y_1 + x_1 + ...x_m,y_1 + y_2 + x_1,...y_1 + y_2 + x_1 + ... + x_m,...y_1 + ... + y_n + x_1 + ... + x_m$ are equal then we are done. The induction proof is nice, but we can also find a solution along these lines (I think EmptySet showed me this). Let $s_i=x_1+\cdots+x_i$ and $t_j=y_1+\cdots+y_j$ ($s_0=t_0=0$). WLOG $s_m>t_n$ (if they're equal we're done). Now consider $s_i+t_j\pmod{s_m}$ for $i=1,2,\ldots,m$ and $j=0,1,\ldots,n$; since $m(n+1)>s_m$, there exist distinct pairs $(i_1,j_1)$ and $(i_2,j_2)$ in this range of indices (WLOG $j_1\ge j_2$) such that $s_{i_2}-s_{i_1} \equiv t_{j_1}-t_{j_2}\pmod{s_m}$. Since $i_1,i_2>0$ and $t_n<s_m$, neither side vanishes (or else $(i_1,i_2)=(j_1,j_2)$). Thus $j_1>j_2$. If $i_2>i_1$, we get $t_{j_1} - t_{j_2} = s_{i_2} - s_{i_1}$; if $i_2<i_1$, we instead have $t_{j_1} - t_{j_2} = (s_m-s_{i_1}) - (s_m-s_{i_2})$. Either way, we're done.

I'll post a solution ,but i'm not 100% sure if it's right.If there's any mistake,please point it out,thank you~ $x_1<x_2<x_3<……<x_m<x_1+x_m<……<x_(m-1)+x_m<x_1+x_(m-1)+x_m<……<\sum x_i$ There are $\sum_{k=1}^n k=\frac{m(m+1)}{2}$ numbers in total and all of which are less than$mn+m$ Similarly,we replace the $x$ with $y$, and we get $\frac{n(n+1)}{2}$ sums in total and all of which are less than $mn+n$ WLOG we assume that $n\geq m$ thus we get $\frac{m^2+n^2+m+n}{2}$sums of $x_i$ or $y_i$ less than $mn+n$ but $\frac{m^2+n^2+m+n}{2}\geq mn+n$（Because it is equilavent to$(n-m)(n-m-1)\geq 0$ which is obvious so there must be two which have the same value. Q.E.D