Robert Gerbicz wrote: For a given positive integer n consider the numbers $ {a}_{k,l}$ where $ k*l\leq n$. There are $ \sum_{k=1}^{n}\lfloor \frac{n}{k}\rfloor$ such pair of $ (k,l)$ indexes. And this is at least $ \sum_{k=1}^{n}(\frac{n}{k}-1)=n*(-1+\sum_{k=1}^{n}\frac{1}{k})$, but it is well known that $ \sum_{k=1}^{n}\frac{1}{k}$ is divergent so for large $ n$ value there are more than $ 8*n$ pair of $ (k,l)$ indexes meaning that there is at least one $ (k,l)$ pair of indexes for that $ k*l\leq n$ and $ a_{k,l}>n$ (because there are $ 8*n$ pairs of $ (k,l)$ for that $ a_{k,l}\leq n$ is true) from this $ a_{k,l}>k*l$ is also true what is needed.