Only 3628800 Cases

We are adding all the integers $1, 2, 3, \dots 10$ at least once in some order, and adding all the $n$ 's of $a_1, a_2, a_3, \dots a_{10}$ in order, for a total of $1 + 2 + 3 + \dots + 10 + 1 + 2 + 3 + \dots + 10 = 90$ , so the sum of all the unit digits of $a_n + n$ must end in $0$ .

However, since all $10$ unit digits of $a_n + n$ must be different, all $10$ unit digits $0, 1, 2, \dots 9$ are represented, but $0 + 1 + 2 + \dots + 9 = 45$ , which ends in $5$ , not $0$ .

Therefore, $\boxed{0}$ possible arrangements are possible.

No math required:

By looking at the digit sum of $a_n$ and $n$ , we expected to have 5 odds and 5 evens (since all 0 to 9 are expected.

Let's look at an initial condition $(1,1),(2,2),...,(9,9)$ , the digit sum has 0 odds and 10 evens.

When swapping the first number between pairs, the number of odds either increase or decrease by 2. (Invariance!)

The invariance of this swapping operation is that the number of odds and evens must change by only multiples of 2.

Since we have 0 odds and 10 evens, there is no such way to change from 0 odds 10 evens to 5 odds 5 evens, which means no such configuration exists!