The desired number must have the maximum different decimal digits,10 digits,from 0 to 9.From condition of maximality,the last two digits are 20,for divisibility with 2 ,and 4,and ,for divisibility for 8,the last three digits are 120.The sum of digits is 45,so the number is divisible with 3.9 and 6.The difficult is divisibility with 7. Check the number 9876543120; for this,delete 0,the last digit,and obtain 213456789,after rollover. The remainders of 1,10,100,1000,10000,100000 is.in order:1,3,2,6,4,5 and further it repeats cyclically Then: 2x1+1x3+3x2+4x6+5x4+6x5+7x1+8x3+9x2=134 equivalent to 1.mod 7 But this sum must be 133.and , for this,after trials changes the order between 5 aqnd Finally ,the desired number is 9875643120

We should aim to keep as many digits as possible. Definitely, the number $N$ cannot contain $0$ as $0 \nmid N$. Then, the number $N$ should not contain $5$, otherwise $N$ cannot contain $2, 4, 6 $ and $8$. Note that $9 \nmid 40 = 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9$ While $9 \mid 36 = 1 + 2 + 3 + 6 + 7 + 8 +9$ Therefore, $N$ shall have maximum $7$ digits. For $N$ to be largest and divisible by $8$, the last three digits shall be $312$. Now Let's try to arrange first four digits $(9, 8, 7, 6)$, thus $N$ shall be largest one and be divisible by $7$. $$7 \nmid 9876312$$$$7 \mid 9867312$$Therefore, the desired natural number is $\boxed {9867312}$.

All numder ,including zero,divide zero.So,my solution is correct

@above that is true, but no other number except zero is divisible by zero.