We see that \[N=(10-1)+(100-1)+\dots+(10^{2019}-1)=\underbrace{11111111\dots1}_{2019}0-2019\]and we know \[11110-2019=9091\]so there are $2019-4+1=\boxed{2016}$ ones in the decimal representation of $N$.

We see that \[N=(10-1)+(100-1)+\dots+(10^{2019}-1)=\underbrace{11111111\dots1}_{2019}0-2019\]and we know \[11110-2019=9091\]so there are $2019-4+1=\boxed{2016}$ ones in the decimal representation of $N$.