If the digits are all different, then the sum of all possible three-digit numbers is $n = 10000a+1000b+100c+10d+e = 12 \cdot 111(a+b+c+d+e)$ Note that $\bmod 9$ we conclude that $9 | a+b+c+d+e$. Let $a+b+c+d+e = 9d$; then $n = 11988d$ When $d = 3$ we have $n = \boxed{ 35964}$, with digit sum $27 = 9d$.