jmc

Since there are 4 band members left over when they line up in rows of 26, we have $20n\equiv 4(\mathrm{mod}26)$. We divide both sides of the congruence by 4, remembering that we have to divide 26 by the greatest common divisor of 4 and 26. The original congruence is equivalent to $$5n\equiv 1(\mathrm{mod}13).$$So we would like to find a multiple of 13 which is one less than a multiple of 5. Noticing that $13\cdot 3$ has a units digit of 9, we identify $(13\cdot 3+1)/5=8$ as the inverse of 5 (mod 13). Multiplying both sides of our congruence by 8 gives $$n\equiv 8(\mathrm{mod}13).$$We have found that $n$ satisfies the conditions given in the problem if $n=8+13k$ for some positive integer $k$ and $20n<1000$. Rewriting the inequality $20n<1000$ as $n<50$, we solve $8+13k<50$ to find that the maximum solution is $k=\lfloor 42/13\rfloor =3$. When $k=3$, the number of band members is $20(8+13(3))=\boxed{940}$.