Let $n$ be any integer greater than $2^{10^6}$. Given any $s\le (\log\_2 n)/1000$ integers $1=q\_1\le q\_2\le \cdots q\_{s-1}\le q\_s=n$. Prove that
$$\min\_\ell\left(\sum\_{i=1}^\ell q\_i\right)\left(\sum\_{i=\ell+1}^s\frac{1}{q\_i}\right)\le \frac{1}{64}$$



Min problem on integers
Let us denote
$$\sigma\_\ell:=\sum\_{i=1}^\ell q\_i\qquad\text{and}\qquad\tau\_\ell:=\sum\_{i=\ell+1}^s\frac{1}{q\_i}.$$
Then
$$\prod\_{\ell=1}^{s-1}\left(\frac{q\_\ell}{q\_{\ell+1}}\cdot\frac{\sigma\_{\ell+1}}{\sigma\_\ell}\cdot\frac{\tau\_{\ell-1}}{\tau\_\ell}\right)=\frac{q\_1}{q\_s}\cdot\frac{\sigma\_{s}}{\sigma\_1}\cdot\frac{\tau\_0}{\tau\_{s-1}}=\sigma\_s\tau\_0>n,$$
hence there exists $\ell\in\{1,2,\dotsc,s-1\}$ such that
$$\frac{q\_\ell}{q\_{\ell+1}}\cdot\frac{\sigma\_{\ell+1}}{\sigma\_\ell}\cdot\frac{\tau\_{\ell-1}}{\tau\_\ell}>n^{1/(s-1)}>n^{1/s}=2^{1000}.$$
On the other hand, the left-hand side equals
$$\frac{q\_\ell}{q\_{\ell+1}}\left(1+\frac{q\_{\ell+1}}{\sigma\_\ell}\right)\left(1+\frac{1}{q\_\ell\tau\_\ell}\right)
=\frac{q\_\ell}{q\_{\ell+1}}+\frac{q\_\ell}{\sigma\_\ell}+\frac{1}{q\_{\ell+1}\tau\_\ell}+\frac{1}{\sigma\_\ell\tau\_\ell},$$
so for the same $\ell$ we have
$$\frac{q\_\ell}{q\_{\ell+1}}+\frac{q\_\ell}{\sigma\_\ell}+\frac{1}{q\_{\ell+1}\tau\_\ell}+\frac{1}{\sigma\_\ell\tau\_\ell}>2^{1000}.$$
The first three terms on the left-hand side do not exceed $1$, hence in fact
$$\sigma\_\ell\tau\_\ell<\frac{1}{2^{1000}-3}.$$


