24

$2018(2019^{39}+2019^{37}+...+2019)+1$ =$(2019-1)(2019^{39}+2019^{37}+...+2019)+1$ =$2019^{40}-2019^{39}+2019^{38}...+2019^{2}-2019+1$ =$1-2019+2019^{2}-2019^{3}...-2019^{39}+2019^{40}$ The above is a Geometric series whose sum evaluates to $(2019^{41}+1)/2020$ From question we know that above is a prime. Let's call the prime $p$. $(2019^{41}+1)/2020$ = p Which implies that $(2019^{41}+1)$ = $p*2020$ = $2^{2}*5*101*p$ $p$ is obviously not 2, 5 or 101. It is much larger. So the above is a valid prime factorisation. We can then get the number of positive factors of $(2019^{41}+1)$ to be $3*2*2*2 = 24$