I'm not sure if trivial solutions for odd n are possible, but they wouldn't be the largest.
The trivial solution for even n isn't always the biggest.
10 9999996699 9999986701 99999834000043899999
18 999999999889625119 999999999580927521 999999999470552640046255074999999999
20 99999999998397393961 99999999998547088359 9999999999694448232002328444969999999999
22 9999999999993257203059 9999999999959141742661 99999999999523989457200275498932599999999999
As you have observed, you cannot have any number ending with a $0$, so your first two digits are fixed.
Division tests for numbers ending with $1,2,5,6$ add no information.
From the division test for $3$, you know that the first two digits must sum to a multiple of $3$.
From the division test for $7$, you know that the number obtained by truncating the last digit and subtracting away twice the last digit from it will give a multiple of $7$. This means that the first two digits have to also be a multiple of $7$.
From these two conditions, the first two digits must be a multiple of $21$, which means they can be $21,42,63,84$.
From the division test for $4$, you know the last two digits must form a number divisible by $4$.
The only one that fits this condition is the first two digits $84$.
So the smallest number is $841$, digit sum $13$.
For rigour, you need to exclude the possibility of the sequence starting with first digit $2$. If this were the case, the final number would end with $9$, and the division test for $9$ demands the sum of the first two digits should also be a multiple of $9$. But this would mean that $63$ is the only possibility, which would conflict with the rule for the last digit $4$ ($34$ is not a multiple of $4$). So we have verified that the above is the only sequence of numbers that meets the criteria.
Best Answer
$x+y + xy=10x + y\implies 9x=xy$
If $x\ne 0$ then $y=9$