The computer program found 2554 integers which met the criteria.

Analytic solution follows.

225 = 3*3*5*5, so solutions will be any number which ends in any of {00,25,50,75} and also has a digital root of 9, while being comprised of only 2 unique digits.

Positive integers below 10^12 have up to 12 digits.

If zero is one of the digits, then leading zeros are allowed and we can say ALL such numbers have 12 digits.

case 1: ends in 00; the other digit can be any from 1 to 9.

1A:  If the other digit is k=1,2,4,5,7, or 8 then there must be 9 of them.  The 12 digit number's first 10 digits are one 0 and 9 k's (10 ways).  10 ways times 6 different k's.  1A = 60

1B:  the other digit is k= 3 or 6.  There must be either 3 or 6 or 9 of them.

Ways = [c(10,3)+c(10,6)+c(10,9)] times 2 = (120+210+10)*2 -->   1B = 680

1C:  the other digit is 9, and there can be any number of 9's (at least 1).  So the possible ways of 10 digits being either 0 or 9 (but not all zeros) is 2^10 - 1    1C = 1023

2: ends in 25. 

12-digit numbers:  To get a digital root of 0 from 12 digits which are m 2's and n 5's, (n,m) can be (2,10), (5,7), (8,4), or (11,1); but the last 2 digits are earmarked as being 25, so the number of 2's (with the rest being 5's) can be (1,4,7,10). The combination of 10 things taken (1,4,7,10) at a time is (10,210,120,1) the sum of which is 341.  So 341 ways of 12 digit numbers ending in 25.

Still ending in 25, but less than 12 digits.  Now we have leading blank digits, but still m 2's and n 5's where m+n < 10 (ignoring the final '25').  The 12-digit numbers have already accounted for (m,n) = (1,9) (4,6) (7,3) and (10,0).

Less than 12-digits gives us (1,0) (3,1) (0,4) (5,2) (2,5)

[note for example (1,0) represents 9 blanks then one 2 then 25, or the integer 225].  So these five groups give combinations combin(1,1), combin(4,3), combin(4,0), combin(7,5), combin(7,2)  The sum of these is 48.

3: ends in 50.  Digital root of first 10 digits must be 4. Eight 5's.  Since other digit is zero, leading zeros are OK so c(10,8) = 45  

4: ends in 75.  Lets consider allowable (m,n) for the number of m 7's and n 5's where m+n is from 0 to 10, not counting the final '75'.  Then check combin(m+n,m)  (0,3) (2,2) (1,7) (4,1) (3,6) (5,5) (6,0) gives:

1,6,8,5,84,252,1  whose sum is 357

The grand sum does work out to the same number, 2554

Ending in 00:  60 + 680 + 1023 = 1763

Ending in 25:  341 + 48        = 389

Ending in 50:  45

Ending in 75:  357