Cute problem but too easy for a 10.4. Solved with p_square, Arwen713 We consider the range $[2n+5-a, 2n+5+a]$, and consider $(i,j)\in \{1,\cdots 2n+4\}^2$ and $i\ne j$ such that $i+j\in [2n+5-a, 2n+5+a]$. Observe that there are atleast $(2a+1)(n+1-\frac{a}{4})+1+\frac{a}{4}$ such pairs. Let the set of all such pairs be $S$. Now, consider $a_i+a_j$ for $(i,j)\in S$. We have $a_i+a_j\in [2n+5-a, 2n+5+a+2)$. Now, we have $(2a+1)(n+1-\frac{a}{4})+1+\frac{a}{4}$ pairs in a range of $2a+2$. Thus, we must have some two within a distance of $\frac{2a+2}{(2a+1)(n+1-\frac{a}{4})+\frac{a}{4}}$. Now, let $a=\lfloor\sqrt{2n}\rfloor$ and we get the desired bound. Now, the only issue is the numbers might be $(a_i+a_j)-(a_j-a_k)=a_i-a_k$. Now, remove all pairs including $k$ and the bound still works. But now, if we run into a similar issue again i.e. a pair $a_l-a_m<\frac{1}{n-\sqrt{\frac{n}{2}}}$, we will have $a_i+a_m-a_k-a_l<\frac{1}{n-\sqrt{\frac{n}{2}}}$ and if $l=i$ or $m=i$, we will have 3 distinct numbers within a distance of $\frac{1}{n-\sqrt{\frac{n}{2}}}$ of each other, which is not possible as $a_{i+2}-a_i>1$.

Nice algebraic problem, proposed by Alexander Kuznetsov

Rg230403 wrote: Cute problem but too easy for a 10.4. Unfortunately it is tough topic for children, only 1.5 contestants solved this problem.

Oh interesting, that changes things. My view on this was mostly due to the fact that the idea felt easy to motivate. The most natural bound to try is with $(a_{1},a_{2n+4}), (a_2,a_{2n+3},\cdots )$ initially and then you can observe that the reason it's not strong enough is the fact that the range of the sums is $2$ when we were only going for pairs $i,j$ with $i+j$ taking one value. So if we take pairs $i+j$ which take a set of $k$ values then the range of the resultant should be $k+1$ which should be a better bound and now its a matter of selecting $k$ which can be motivated by the problem statement.

Rg230403 wrote: Oh interesting, that changes things. My view on this was mostly due to the fact that the idea felt easy to motivate. The most natural bound to try is with $(a_{1},a_{2n+4}), (a_2,a_{2n+3},\cdots )$ initially and then you can observe that the reason it's not strong enough is the fact that the range of the sums is $2$ when we were only going for pairs $i,j$ with $i+j$ taking one value. So if we take pairs $i+j$ which take a set of $k$ values then the range of the resultant should be $k+1$ which should be a better bound and now its a matter of selecting $k$ which can be motivated by the problem statement. Yes I agree that this way is very natural. Now I suggest you to prove bound $\frac{1}{n-C}$ for some positive constant $C$ and for big enough positive integer $n$.