Proof:Lemma:\[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_n} \] The proof of Lemma:When $n=1$,it's easy to see it's correct. Suppose it's correct for $n-1(n\geq2)$, then \[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_{n-1}}-\frac{1}{r_n}=\frac{1}{r_n-1}-\frac{1}{r_n}=\frac{1}{(r_n-1)r_n}=\frac{1}{r_1r_2\ldots r_n} \] That means the lemma is correct. We only need to show that $\forall a_1\leq a_2\leq\ldots\leq a_n\in\mathbb{N},\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}<1$, then we must have \[ \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\leq\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n} \] We prove it by induction: When $n=1$,$\frac{1}{a_1}<1$,$a_1\geq2=r_1$.So $\frac{1}{a_1}\leq\frac{1}{r_1}$. Suppose the conclusion is correct for all positive integer smaller than $n$,and there exist $a_1,a_2,\ldots,a_n$ such that \[ \frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n}<\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}<1 \eqno(*) \] By Lemma\[ 1-(\frac{1}{r_1}+\frac{1}{r_2}+\ldots+\frac{1}{r_n})=\frac{1}{r_1r_2\ldots r_n} \] And it's obvious that \[ 1-(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n})\leq\frac{1}{a_1a_2\ldots a_n} \] Because $\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}=\frac{x}{a_1a_2\ldots a_n}<1,x\in\mathbb{Z}$ So $x<a_1a_2\ldots a_n$,and $x\leq a_1a_2\ldots a_n-1$,so \[ 1-(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n})=1-\frac{x}{a_1a_2\ldots a_n}\leq\frac{1}{a_1a_2\ldots a_n} \] So by (*) \[ a_1a_2\ldots a_n\geq r_1r_2\ldots r_n \eqno(**) \] And by the transform of Abel \[ \begin{eqnarray*} \sum_{i=1}^n\frac{a_i}{r_i}&=&a_n\sum_{i=1}^n\frac{1}{r_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{r_i})(a_k-a_{k+1})\\ &<&a_n\sum_{i=1}^n\frac{1}{a_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{a_i})(a_k-a_{k+1})\\ &=&\sum_{i=1}^n\frac{a_i}{a_i}=n \]\end{eqnarray*} By AM-GM \[ n>\sum_{i=1}^n\frac{a_i}{r_i}\geq n\sqrt[n]{\frac{a_1a_2\ldots a_n}{r_1r_2\ldots r_n}} \] So \[ a_1a_2\ldots a_n<r_1r_2\ldots r_n \] And it's a contradiction with (**). So there doesn't exist $a_1,a_2,\ldots,a_n$ satisfied (*),that means the conclusion is correct.

isn't it obvious? we take the smallest integer such that it doesn't exceed or equal 1..and it's easy to show that it always works..?

srulikbd wrote: isn't it obvious? we take the smallest integer such that it doesn't exceed or equal 1..and it's easy to show that it always works..? well, that not necessary give the largest sum as a whole

I suppose the messy code is: $\sum_{i=1}^n\frac{a_i}{r_i}=a_n\sum_{i=1}^n\frac{1}{r_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{r_i})(a_k-a_{k+1})<a_n\sum_{i=1}^n\frac{1}{a_i}+\sum_{k=1}^{n-1}(\sum_{i=1}^k\frac{1}{a_i})(a_k-a_{k+1})=\sum_{i=1}^n\frac{a_i}{a_i}=n$

Pretty much as a problem from Bulgaria RMM TST 2023.