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Let's brute force this; start with

9 numbers ending on 1, e.g. 1, 11, 21, 31, 41, 51, 61, 71 and 81. Now, we cannot add another number ending in 1. But we can add a few ending in 2; each sum we can make has $a$ '2s' and $10-a$ '1s', so its last digit is $a$. As long as we have 9 numbers ending on 2, $1 \le a \ge 9$ and the sum won't be divisible by 10. We cannot add more numbers; if we add a single car number ending in $x$, we can take $b$ car numbers ending on 2, $9-b$ ending on 1 and $x$ to produce a sum which is equal to $9 + b + x \pmod {10}$, so just choose $b = 1-x$ (if $x=0,1$) or $b = 11-x$.

So the final answer is

18.

Let's brute force this; start with

9 numbers ending on 1, e.g. 1, 11, 21, 31, 41, 51, 61, 71 and 81. Now, we cannot add another number ending in 1. But we can add a few ending in 2; each sum we can make has $a$ '2s' and $10-a$ '1s', so its last digit is $a$. As long as we have 9 numbers ending on 2, $1 \le a \ge 9$ and the sum won't be divisible by 10. We cannot add more numbers; if we add a single car number ending in $x$, we can take $b$ car numbers ending on 2, $9-b$ ending on 1 and $x$ to produce a sum which is equal to $9 + b + x \pmod {10}$, so just choose $b = 1-x$ (if $x=0,1$) or $b = 11-x$.

So my answer is

18.

Let's brute force this; start with

9 numbers ending on 1, e.g. 1, 11, 21, 31, 41, 51, 61, 71 and 81. Now, we cannot add another number ending in 1. But we can add a few ending in 2; each sum we can make has $a$ '1s' and $10-a$ '2s', so its last digit is $10-a$. So as long as we have 9 numbers ending on 2, $1 \le a \ge 9$ and the sum won't be divisible by 10. We cannot add more numbers, as we can make any end digit with 9 cars which end in either 1 or 2.

So the final answer is

18.

Let's brute force this; start with

9 numbers ending on 1, e.g. 1, 11, 21, 31, 41, 51, 61, 71 and 81. Now, we cannot add another number ending in 1. But we can add a few ending in 2; each sum we can make has $a$ '2s' and $10-a$ '1s', so its last digit is $a$. As long as we have 9 numbers ending on 2, $1 \le a \ge 9$ and the sum won't be divisible by 10. We cannot add more numbers; if we add a single car number ending in $x$, we can take $b$ car numbers ending on 2, $9-b$ ending on 1 and $x$ to produce a sum which is equal to $9 + b + x \pmod {10}$, so just choose $b = 1-x$ (if $x=0,1$) or $b = 11-x$.

So the final answer is

18.

I think the answer is

9, and possible numbers are 1, 11, 21, 31, 41, 51, 61, 71, 81. If you take the sum of $1 \le n \le 9$ numbers, the last digit of the sum is $n$ so the sum is not divisible by 10.

Let's brute force this; start with

9 numbers ending on 1, e.g. 1, 11, 21, 31, 41, 51, 61, 71 and 81. Now, we cannot add another number ending in 1. But we can add a few ending in 2; each sum we can make has $a$ '1s' and $10-a$ '2s', so its last digit is $10-a$. So as long as we have 9 numbers ending on 2, $1 \le a \ge 9$ and the sum won't be divisible by 10. We cannot add more numbers, as we can make any end digit with 9 cars which end in either 1 or 2.

So the final answer is

18.