the number $ 9999 $ verifies the statement. so we need to find (if exists) a number with at least $ 5 $ digits. let $ n=a_{1}....a_{m}abc $ with $ a_{i},a,b,c $ digits and $ c>0 $. then $ a_{1}.....a_{m}c $ will divide $ n $ so this number will be also a divisor of the modulus of $ abc-100c $. this number has at most three digits do $ m \leq 2 $. let be $ n=abcde $ with $ a,b,c,d,e $ digits and $ e>0 $. then $ n $ is divisible by $ ade $ and by $ abe $. then $ abe $ will divide $ cde-100e $. the modulus of this number is $ <900 $ so $ a \leq 8 $. let be $ a=8 $. then $ c=9,e=1,b=d $ or $ c=0,1 $. for obtain a maximal number we have to take the first case. now is easy to see that $ n\leq 89991 $. so the maximum is $ 89991 $.

Let $n = \overline {{a_1}...{a_{k - 2}}{a_{k - 1}}{a_k}} $, $k \geqslant 4$. Then $m = \overline {{a_1}...{a_{k - 2}}} $ satisfy $m\left| {n - 100m} \right.$, so $m\left| {10{a_{k - 1}} + {a_k}} \right.$, hence $m \leqslant 99$, and $n \leqslant 9999$. So $n = 9999$ is maximum.

did you see what I wrote? please read again the problem. you cannot do that.