Answer: 59

Solution:

Write $a_k = \frac{m}{n}$ in lowest terms and let $d = n - m,$ so $a_j$ has the form $\frac{t}{t+1}$ exactly when $d = 1.$ One step sends $(m, n)$ to $(m + 18,\, n + 19)$ and then cancels $g = \gcd(m + 18, n + 19).$ Two facts control everything. First, $I = 19m - 18n$ satisfies $19(m + 18) - 18(n + 19) = I,$ so $I$ is unchanged by the shift and divided by $g$ upon cancellation. Second, since $I = 19(m + 18) - 18(n + 19)$ $= (m + 18) - 18(d + 1),$ a number divides both $m + 18$ and $n + 19$ exactly when it divides both $d + 1$ and $I;$ hence $g = \gcd(d + 1,\, I),$ and after cancelling, the new difference is $\frac{d + 1}{g}.$

Initially $I = 19 \cdot 2020 - 18 \cdot 2021 = 2002$ $= 2 \cdot 7 \cdot 11 \cdot 13$ and $d = 1,$ so $j = 1$ works. The next step has $d + 1 = 2,$ $g = 2:$ $a_2 = \frac{2038}{2040} = \frac{1019}{1020},$ so $j = 2$ works and $I = 1001 = 7 \cdot 11 \cdot 13.$ From there $d$ climbs $1, 2, 3, \ldots$ until $d + 1$ shares a factor with $I:$ at $d + 1 = 7$ we get $a_8 = \frac{161}{162}$ (cancel $7,$ now $I = 143$); then at $d + 1 = 11,$ $a_{18} = \frac{31}{32}$ (cancel $11,$ now $I = 13$); then at $d + 1 = 13,$ $a_{30} = \frac{19}{20}$ (cancel $13,$ now $I = 1$). Each cancellation used $g = d + 1,$ so $d$ returned to $1$ at $j = 8,$ $18,$ and $30.$

Once $I = 1,$ no further cancellation is possible, so $d$ increases forever and never equals $1$ again. The valid indices are $j = 1, 2, 8, 18, 30,$ with sum $1 + 2 + 8 + 18 + 30 = 59.$