0.222...*9=2

0.2525252...*99=25

0.2018201820...*9999=2018

0.2018201820...*99999999=20182018

0.2018201820...* (9999... n ....9999)=(2018... n ... 2018)

2019=673*3 (no other divisors) => p will be multiple of 2019 if it is multiple of 673 and is multiple of 3

1/673=0,001485884... as a not-regular fraction the decimal part is repeated after 672 digits or less. Concretely for number 673 the decimal part has a period of 224 digits. (Big online numbers calculator for this)

As we have seen in the beginning of the post: 

0,001485884... (follows digits since 224, and repeated infinite times) can be express as:

001485884.../9999...-224 nines-...999 (where both numbers are integers).

Then I can choose number p (=2018... n ....2018) with n=56 groups (2018) and with a number of digits 56*4=224 digits. 

p/673=p*0,01485884...

=p*(001485884.../9999...-56 groups-..9999)

=p*001485884...*(0,20282028.../p)

So:

p=673*001485884...*0,20282028....

This is an integer as p is an integer an, we can see it, is a multiple of 673. In addition 673 is prime so 001485884...*0,20282028.... is an integer.

Let see when p is not just multiple of 673 but of 2019 (673*3). 

Here we will need to work with three full decimal periods of 1/673. 

(0.001485884....001485884....001485884)...001485...=

=001485... 001485... 001485 -integer with 224*3 digits- *

* 9999... 224*3 nines ...9999

See that now: =001485... 001485... 001485 -integer with 224*3 digits- is obviously a multiple of 3.

Repeating what we did before, the new p composed of groups of number 2018 will be:

p=673*001485... 001485... 001485...*0,20282028...

The product of the last two terms will be an integer again. Then:

p will be now multiple of 2019, will have now 224*3 digits=672 digits, so that it will be composed by 168 groups of 2018.