${m=2001k,k=S(2001*k)<9*[lg(2001k)+1}\to k<50$. Therefore $S(m)\le 9+8+4+9=30$. Therefore $S(m)\le 5+8+2+9=26$. It gives $S(m)\le 21$. It gives $S(m)\le 18$. If $k<10$, then $S(2001k)>k$. If $10<k<18$, then $S(2001*k)<k$. Work only $k=18$.

we have the classical theorem:$S(m)<9lgm$ hence $18089lgm>m,2001\le m\le 89580$ mod3,9,it's easy to prove $27|m$ hence $S(m)=9,18,27,36,45$ then it's easy to confirm $m=36018$ is the only solution.