I think the answer is
The number cannot contain a $5$, since deleting the $5$ means the result cannot end in $5$ or $0$.
If the number contains a $9$, then the sum of its digits is divisible by $9$. $1+2+3+4+6+7+8+9 = 40$ so if we wish the number to contain a $9$, it can have at most $7$ digits. If we want to pursue this path, we would have to exclude $4$ to guarantee divisibility by $9$. This leaves us to construct a $7$-digit number from $1,2,3,6,7,8,9$ with the given property.
The number must end in an even digit to ensure divisibility by each even number and also, the second-last digit must be even to ensure divisibility by the last digit.
If the last two digits are $26$ or $62$ then we cannot get divisibility by $8$. Hence, $8$ must be one of the last two digits.
Now suppose we wish to make the first two digits $97$ to get the maximum number possible. If we require $2$ to be the other one of the last two digits then the fourth and fifth digits must either be $31$ or $63$ to guarantee divisibility by $8$ when $8$ is deleted.
Therefore, the numbers to test in this branch are $9763182, 9763128, 9716382$ and $9716328$. Only the last number passes divisibility by $7$ which is $9716328$ so this is our best answer is this branch.
Hence, we must have that the number begins with the digits $97$ and several of our unexplored branches can be discounted because numbers there will be strictly less than $9700000$.
There remains one branch to explore which is when $6$ is the other of the last two digits (there could be a possible larger solution here). In this case, the fourth and fifth digits must be $21$ or $13$ to ensure divisibility by $8$ when $8$ is deleted.
This leaves four numbers to test: $9732168, 9732186, 9721368, 9721386$.
Again, there is just one candidate which satisfies the divisibility by $7$ rule and that is $9721368$. This is bigger than our previous candidate and hence is the maximum.