How many two-digit numbers have the sum of the digits of the sum of digits equal to 1?

So, at first glance, this is the number 10, where the sum of digits = 1 + 0 = 1, but given that, by the definition of many properties of divisibility, the sum of digits is determined to a single digit, then in the following numbers the sum of digits = 1: 19 (1+ 9 = 10,1 + 0 = 1), 28 (2 + 8 = 10, 1 + 0 = 1), 37 (3 + 7 = 10 ..), 46, 55, 64, .73,82,91, and as mentioned it is 10. As a result, we get 10 numbers with the named property - sum of digits = 1.

The sum of digits of the sum of digits of a two-digit number (let's call it "double sum of digits") is equal to 1 only in one case, if the first sum of digits is 10. So these are numbers like: 19,28, 37 ... and so on ... 82, 91. There are only 9 such numbers, there is also a ten itself, in total, 10 two-digit numbers have a double sum of digits equal to 1.