imc

Notice that $1000$ is $3E8$ when converted to hexadecimal ($3\cdot 16^2+14\cdot 16^1+8\cdot 16^0$). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal. The first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0-9$, giving $4\cdot 10\cdot 10=400$ combinations. However, this includes $000,$ so this number must be diminished by $1.$ Therefore, there are $399$ valid $n$ corresponding to those $399$ positive integers less than $1000$ that consist of only numeric digits. (Notice that $399$ is the least hexadecimal number using only decimal digits before $3E8$.) Therefore, our answer is $3+9+9=\boxed{21}$