( @mavropnema or someone else who knows: Juniorsection: degree $\le 10$? ) Solution. Just use induction to the following arguments. Notations. Let $R_i$ be row $i$, and $K_i$ be column $i$. Let $\sigma(R_i)$ be the sum of elements of row $i$, and $\sigma(K_i)$ be the sum of elements of column $i$. $A<(R_i,K_i)<B$ means the minimum of $\sigma(R_i), \sigma(K_i)$ is larger than $A$, and the maximum of the two is smaller than $B.$ The $n^2$ distinct numbers are $b_1<b_2<\cdots <b_{n^2}$. We can put $n^2$ numbers in such a way that $\sigma(R_1)<\sigma(K_1) < (R_2,K_2)<\cdots < (R_n,K_n)$ and $b_{n^2} \in R_n \cap K_n$ for each $n \ge 2$. The $n^2$ numbers are $b_1<b_2<\cdots <b_{n^2}$. $\begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \\ \end{bmatrix}$ works for $n=2.$ Now to go to $(n+1)^2$, place $b_{n^2+i}$ next to the $i^{th}$ smallest row/column for $i \in \{1,2\cdots , 2n\}$, and $b_{(n+1)^2}$ in the new corner. So $R_1 \cap K_{n+1}= b_{n^2+1}$, $R_{n+1} \cap K_1= b_{n^2+2}$, and so on. Then $\sigma(A_i)+b_i<\sigma(A_{i+1})+b_i$ where $A_i$ are the rows and columns in order, for $i$ until $2n+1.$ If $\sigma(R_{n+1}) \neq \sigma(K_{n+1})$ we are done. Else change $b_{n^2}$ with $b_{n^2+1}$ The statement is now true: * $\sigma(R_1)$ became yet smaller, but was already the smallest one, so no equality with an other $\sigma(A_i)$ after the operation. * $\sigma(R_n),\sigma(K_n)$ becomes bigger and keeps bigger than the rows,columns with lower indices. * $\sigma (R_{n+1})>\sigma(K_{n+1}).$ * $(R_n,K_n) < (R_{n+1},K_{n+1})$ proof : $K_{n+1}>b_{n^2}+(n-1)b_{n^2+2}+b_{(n+1)^2}$ $>b_{n^2+1}+(n-1)b_{n^2-1}+b_{n^2+2n}>(k_n,R_n)$ Hence with complete induction it goes for all $N$ as we checked the last stap for the $4$ changed values.$

SCP, I have cleaned your write-up (there where a number of typos and obscure statements), but I think there is a flaw, at the very end. If we end up with $\sigma(R_{n+1}) = \sigma(K_{n+1})$, and you interchange $b_{n^2}$ with $b_{n^2+1}$, indeed $\sigma(R_1)$ gets even smaller, and $\sigma(R_{n+1})$ becomes not equal to $\sigma(K_{n+1})$, but what makes you write $(R_n, K_n) < (R_{n+1}, K_{n+1})$ ? Both $\sigma(R_{n})$ and $\sigma(K_{n})$ have increased, by $b_{n^2+1} - b_{n^2}$, while $\sigma(K_{n+1})$ has decreased, by $b_{n^2+1} - b_{n^2}$, so it may become equal to one of the above. Yes, I think Juniors league means grade at most $10$.

SCP wrote: * $\sigma (R_{n+1})>\sigma(K_{n+1}).$ * $(R_n,K_n) < (R_{n+1},K_{n+1})$ proof : $K_{n+1}>b_{n^2}+(n-1)b_{n^2+2}+b_{(n+1)^2}$ $>b_{n^2+1}+(n-1)b_{n^2-1}+b_{n^2+2n}>(k_n,R_n)$. Now that you've edited it, a minor flaw still remains. First, you can't say $\sigma (R_{n+1})>\sigma(K_{n+1})$, because you doný know how the $b_{n^2+i}$ are distributed within them; for each $1<i\leq n$, you don't know if $\sigma (R_{i})>\sigma(K_{i})$ or $\sigma (R_{i})<\sigma(K_{i})$. Therefore you should just say $(K_{n+1},R_{n+1})>$ $b_{n^2}+(n-1)b_{n^2+2}+b_{(n+1)^2} >$ $ b_{n^2+1}+(n-1)b_{n^2-1}+b_{n^2+2n}>$ $(K_n,R_n)$, true since it can be written $(b_{n^2} - b_{n^2-1}) +(b_{n^2+2} - b_{n_2+1}) + (n-2)(b_{n^2+2} - b_{n^2-1}) +(b_{(n+1)^2} - b_{n^2+2n}) > 0$. Now everything seems ok.

So, of course, the specification $N>10$ is rubbish; more importantly, the problem remains true for arbitrary distinct real numbers (not necessarily integer, neither positive). The official solution itself makes use of a classical combinatorial Lemma, which has a simple, elementary proof for integers; it can however be extended to real numbers via a number of methods (one being Linear Algebra). As a rule, it is seldom - if ever - the case for such problems to not admit an extension from integers to real numbers, albeit occasionally requiring higher techniques.

mavropnevma wrote: SCP, Yes, I think Juniors league means grade at most $10$. No, Junior league means grade at most $ 9$!.

Lemma: If we have $2n+1$ positive integers such that however we choose one of them, the others could be distributed into two groups of $n$ numbers such that the sum of the numbers of one group is equal to the sum of the numbers of the other group, then all the numbers are equal. I think this is a well-known problem and I'm not going to waste time with its proof. We go on induction. Trivial case $n=2$ is clear. Suppose now the statement is true for $n=2,3,...,k$, and let's prove it for $n=k+1$. Choose the greatest $2k+1$ numbers from the $(k+1)^2$ given, and for the other $k^2$ numbers apply the induction hypothesis for the top-left $k\times k$ table. As a consequence of the Lemma, we could put one of the $2k+1$ numbers into the bottom-right cell such that however we distribute the other $2k$ numbers, the sum of the $k+1^{th}$ row won't be equal to the sum of the $k+1^{th}$ column. Finally, if we associate the greatest remained number to the line (row or column) with the greatest sum, and so on, our induction will be complete. Remark. The statement remains true for arbitrary distinct real numbers because our Lemma could be easy extended on $\Bbb{Q}$, then using Hamel Basis ($\Bbb{R} /\Bbb{Q}$ is a vectorial space) on $\Bbb{R}$.