The integers are 20,39,58,77 and 96.
EXPLANATION:
Method I (Given by Richard):
Richard's method is provided in this location.

Method II:
Let, f(x) = 10^x – x and g(x) = 99…….99( x2 times).
Then, f(N) = 100*g(N) + u(N); where u(N) = 100 – N.
We proceed to analyse all the multiples of 170 ( including itself), satisfying conditions of the problem
Now, 170 = 18 x 9 +8, giving N= 18+2 = 20, since s.o.d.(100 20) = s.o.d.(80) =8
None of the other quotients and remainders obtained by dividing 170 by 9 would yield a valid N. For example, 170 = 9*17 + 17, but s.o.d.(81) = 9 which is not euqal to 17.
Also, 340 =37 x 9 +7, giving N= 37+2 = 39, since s.o.d.(100 39) = s.o.d.(61) =7
And also, 510 =56 x 9 +6, giving N= 56+2 = 58, since s.o.d.(100  58) = s.o.d.(42) =6
And also, 680 =75 x 9 +5, giving N= 75+2 = 77, since s.o.d.(100 77) = s.o.d.(23) =5
And also 850 = 94 x 9 +4,giving N= 94+2 = 96, since s.o.d.(100 96) = s.o.d.(4) =4
All the other multiples of 170 equal to or greater than 1020 will always yield three digit positive integers. This is a contradiction.
It can easily be verified that None of the other quotients and remainders obtained by dividing 170*p ( for p= 2,3,4,5) by 9 would yield a valid N.
Hence, the integers are 20,39,58,77 and 96.
 
Method III:
Let, f(x) = 10^x – x and g(x) = 99…….99( x2 times).
Then, f(N) = 100*g(N) + u(N); where u(N) = 100 – N.
Let, the minimum value of N, for 10< = N < = 99, be given by N= r.
Then, f(r) = 100*g(r) + 10*s +t; where 10*s + t = 100 –r.
So, s.o.d.(f(r)) = 9(r2) + s+ t.
Also, f(r+19) = 100*g( r+19) + 10(s2)+(t+1);
giving, s.o.d.(f(r+19)) = 9(r2)+ 9*19 + s+t1 = s.o.d.(f(r)) + 170.
Hence, by induction s.o.d.(f(r + 19p)) = s.o.d.(f(r)) + 170*p, where p is a positive integer, whereby r+19p<=99(#)
Now, there can be at most 5 distinct values for p. Accordingly, the last digit of (100r) is any one of 0, 1, 2, 3, 4, 5 and so, (#) will be satisfied only when the last digit of r is any one of 5,6,7,8,9,0.
Hence, s.o.d.(f(r + 19*p)) is divisible by 170 iff s.o.d.(f(r)) is divisible by 170, whenever r+19*p <=99, for p = 0,1,2,, and the last digit of r is any one of 5,6,7,8,9,0.
But, 170 = 9*18 +8.
Since, 10(108) +0 = 20 = 18+2; it follows that r=20, giving N = 20+19*p.
Accordingly, N = 20, 39, 58, 77 and 96 are the only two digit integers satisfying all the conditions of the problem under reference.
