I solved this problem by examining how each potential digit changes the value of SOD(m) to SOD(2m)

Let d be a digit in m. If d=1 then SOD(2m) increases by 1 from SOD(m). Similarily for d=2 to 4, the increase is 2 to 4. If d=5 then there is a decrease of 4. Similarily d=6,7,8 have decreases of 3,2,1. if D=0 or 9 then there is no change from SOD(m) to SOD(2m).

Since SOD(m) and SOD(2m) are close, most of the digits will be 9's to get the SOD(m) up to 100 without changing SOD(2m) too much. Ten 9's will make up 90 of each SOD. Adding a 2 and a pair of 4s gives m=**2,449,999,999,999**. SOD(2,449,999,999,999) = 100 and SOD(4,899,999,999,998)=110