Write $$f(n)=f_{0}(n)+\cdots+f_{9}(n)$$ where $f_{d}(n)=1$ if $n$ contains the digit $d$ and 0 otherwise. The sum of $f_{d}(n)$ over all 2019-digit positive integers $n$ is just the number of 2019-digit positive integers that contain the digit $d$. For $1 \leq d \leq 9$, $$\sum_{n} f_{d}(n)=9 \cdot 10^{2018}-8 \cdot 9^{2018}$$ Also, $$\sum_{n} f_{0}(n)=9 \cdot 10^{2018}-9^{2019}$$ Summing over all possible values of $d$, we compute $$\sum_{n} f(n)=\sum_{d=0}^{9} \sum_{n} f_{d}(n)=9\left(9 \cdot 10^{2018}-8 \cdot 9^{2018}\right)+9 \cdot 10^{2018}-9^{2019}=9\left(10^{2019}-9^{2019}\right)$$