Let $M$ be the set of all positive integers that do not contain the digit $9$ (base $10$). If $x_1, \ldots , x_n$ are arbitrary but distinct elements in $M$, prove that
\[\sum_{j=1}^n \frac{1}{x_j} < 80 .\]
It is a very generous easy estimation.
Between $0$ and $10^k-1$, there are $9^k$ numbers without a digit $9$.
So between $10^k$ and $10^{k+1}-1$, there are $9^{k+1}-9^k=8\cdot 9^k$ numbers without a digit $9$.
Thus \[ \sum_{x\in M}\frac{1}{x}< \sum_{k=0}^{\infty}\frac{8\cdot 9^k}{10^k}=\frac{8}{1-\frac 9{10}}=80 . \]