2015 AMC 10A Problem 18

In hexadecimal, $1000=3E8_{16}.$ Pad each shorter representation to three digits with leading zeros. Every numeric-only string from $000_{16}$ through $399_{16}$ is at most $3E8_{16},$ and these are all the possibilities. The first digit has $4$ choices, and each of the other two digits has $10$ choices. This gives $$4\cdot10\cdot10=400$$ strings, including $000_{16}.$ Excluding zero leaves $n=399.$ The sum of the digits in $399$ is $21.$ Thus, E is the correct answer.