Squeeze your brain!

Here is an outline of a solution but in order to keep it reasonably brief I will leave some of the case checking to the reader:

$10 \vert \overline{abcdefghij} \Rightarrow 10 \vert j \Rightarrow j=0 \\ 5 \vert \overline{abcde} \Rightarrow 5 \vert e \Rightarrow e=5 \quad \quad \color{#3D99F6}{\text{As } e \neq 0}$

Note due to the divisibility by $2,4,6,8$ we have that $b,d,f,h$ are all even so $a,c,g,i$ are all odd.

$4 \vert \overline{abcd} \Rightarrow 4 \vert \overline{cd}$

A bit of case checking shows $d=2 \: or 6$ as $c$ is odd.

$6 \vert \overline{abcd5f} \: \text{and} \: 3 \vert \overline{abc} \Rightarrow 3 \vert \overline{d5f}$

Checking cases for $d,f$ gives $d=2,f=8 \: or \: d=6,f=4$ .

Case 1: $d=2,f=8$

$8 \vert \overline{abcd58gh} \Rightarrow 8 \vert \overline{8gh} \Rightarrow 8 \vert \overline{gh}$

As $g$ is odd and $h$ is even checking cases we get $g=1,5,9$ and $h=6$ using numbers we have already assigned to eliminate.

$6 \vert \overline{abc258} \: \text{and} \: 9 \vert \overline{abc258g6i} \Rightarrow 3 \vert \overline{g6i} \Rightarrow 3 \vert (g+i)$

Case checking again shows $g=9,i=3$ is the only possible solution. We also have $b=4$ as it is the only remaining even number.

$7 \vert \overline{a4c2589}$

Checking the two combinations of $1,7$ for $a,c$ we see this yields no solutions.

Case 2: $d=6,f=4$

$8 \vert \overline{4gh} \Rightarrow 8 \vert \overline{gh}$

Checking cases gives $h=2$ , $g=3,7$ :

Case 2.1 $g=3$

$3 \vert \overline{32i} \Rightarrow i=1 \pmod{3}$

As $i$ is odd we have two solutions $i=1,7$ checking the resulting combinations of $a,c$ for divisibility by $7$ shows there are no solutions.

Case 2.2 $g=7$

$3 \vert \overline{72i} \Rightarrow i=0 \pmod{3}$

As $i$ is odd we get $i=3,9$ checking resulting combinations of $a,c$ as above shows $i=9,a=3,c=1$ is the only working triplet so our solution is:

$\color{#20A900}{\boxed{\boxed{3816547290}}}$

Note: I have omitted a lot of case checking in this solution purely to try to keep it to a manageable size (it's already too long).

Python:

import itertools

def test(numstring):
    for i in range(0,10):
        if int(numstring[0:i+1:])%(i+1)!=0:
            return False
    return True

alist = [str(i) for i in range(0,10)]
a=list(itertools.permutations(alist))
for tuples in a:
    seed=""
    for a in tuples:
        seed+=a
    if seed[0]!="0" and test(seed):
        print (seed)

There were too many cases for my liking when I tried to abstain from using brute force, so I resorted to this python program.