Solution:

The answer is (C). Observe that, given a number with decimal representation $abcd$, if we fix $a \neq 0$, we have $10$ choices for each of the digits $b, c, d$: among these $10$ choices, exactly one is "1", just as exactly one is "0". Notice moreover that the same reasoning can be carried out with a three-digit number $abc$ and a two-digit number $ab$.

This allows us to say that, if $a \neq 1$, the number of "1"s and the number of "0"s between $a000$ and $a999$ is the same; if instead $a=1$, we will have $1000$ "1"s (coming from the digit $a$ written a thousand times) more than the "0"s.

Similarly, if $a \neq 1$, the number of "1"s and "0"s between $a00$ and $a99$ and between $a0$ and $a9$ is the same. If instead $a=1$ we have $100$ more "1"s between $100$ and $199$ and $10$ more "1"s between $10$ and $19$.

Let us call $\Delta N$ the difference between the number of "1"s and the number of "0"s in the set of numbers $N$. We can then state that:

- Calling $N_1$ the set of numbers between $1$ and $9$ we have $\Delta N_1=1$;
- calling $N_2$ the set of numbers between $10$ and $19$ we have $\Delta N_2=10$;
- calling $N_3$ the set of numbers between $20$ and $99$ we have $\Delta N_3=0$;
- calling $N_4$ the set of numbers between $100$ and $199$ we have $\Delta N_4=100$;
- calling $N_5$ the set of numbers between $200$ and $999$ we have $\Delta N_5=0$;
- calling $N_6$ the set of numbers between $1000$ and $1999$ we have $\Delta N_6=1000$.

It only remains to count $\Delta N_7$ with $N_7$ the set of numbers between $2000$ and $2020$, but it is easy to verify that $\Delta N_7=-22$. Hence, if $N$ is the set of all numbers between $1$ and $2020$, $\Delta N=\sum_{i=1}^{7} \Delta N_i=1111-22=1089$.