China Southeastern Mathematical Olympiad

Take $A=\{1,2,3,...,35\}$, then for any $a,b\in A$, $$a-b\le 34,a+b\le 34+35=69.$$ In the following, we show that $1\le n\le 69$. Let $A=\{a_1,a_2,\dots ,a_{35}\}$, without loss of generality, suppose that $a_1<a_2<\cdots <a_{35}$.

i. If $1\le n\le 19$, by $$1\le a_1<a_2<\cdots <a_{35}\le 50,$$ $$2\le a_1+n<a_2+n<\cdots <a_{35}+n\le 50+19=69,$$ and by Dirichlet's Drawer Theorem, there exist $1\le i,j\le 35$ ($i\ne j$) such that $a_i+n=a_j$, that is, $a_i-a_j=n$.

ii. If $51\le n\le 69$, by $$1\le a_1<a_2<\cdots <a_{35}\le 50,$$ $$1\le n-a_{35}<n-a_{34}<\cdots <n-a_1\le 68,$$ and by Dirichlet's Drawer Theorem, there exist at least $1\le i,j\le 35$ ($i\ne j$) such that $n-a_i=a_j$, that is, $a_i+a_j=n$.

iii. If $20\le n\le 24$, since $$50-(2n+1)+1=50-2n\le 50-40=10,$$ we see that there are at least 25 elements in $a_1,a_2,\dots ,a_{35}$ that belong to $[1,2n]$. There are at most 24 elements in $\{1,n+1\},\{2,n+2\},\dots ,\{n,2n\}$ such that $\{a_i,a_j\}=\{i,n+i\}$. Hence, $a_j-a_i=n$.

iv. If $25\le n\le 34$, since $\{1,n+1\},\{2,n+2\},\dots ,\{n,2n\}$ have at most 34 elements, by Dirichlet Drawer Theorem, there exist $1\le i,j\le 35$ ($i\ne j$) such that $a_i=i,a_j=n+i$, that is $a_j-a_i=n$.

v. If $n=35$, there are 33 elements $\{1,34\},\{2,33\},\dots ,\{17,18\},\{35\},\{36\},\dots ,\{50\}$. Hence, there exist $1\le i,j\le 35$ ($i\ne j$) such that $a_i+a_j=35$.

vi. If $36\le n\le 50$, if $n=2k+1,\{1,2k\},\{2,2k-1\},\dots ,\{k,k+1\},\{2k+1\},\dots ,\{50\}$; if $18\le k\le 20,50-(2k+1)+1=50-2k\le 50-36=14$; if $21\le k\le 24$, $50-(2k+1)+1=50-2k\le 50-42=8$, there exist $1\le i,j\le 35$ ($i\ne j$) such that $a_i+a_j=2k+1=n$. If $n=2k$, $\{1,2k-1\}$, $\{2,2k-2\}$, ..., $\{k-1,k+1\}$, $\{k\}$, $\{2k\}$, $\{2k+1\}$, ..., $\{50\}$; if $18\le k\le 19$, $50-(2k+1)+3\le 16k-1\le 19-1=18$; if $20\le k\le 23$, $50-(2k+1)+3\le 50-2k+2\le 12k-1$ $\le 23-1=22$; if $24\le k\le 25$, $50-(2k+1)+3\le 50-2k+2\le 4k-1\le$ $25-1=24$, there exist $1\le i,j\le 35$ ($i\ne j$) such that $a_i+a_j=2k$. $\square$