Let's consider a two digit number ab.Its general form will be $10\times a+1\times b$.
If we reverse the digits, its general form will be $10\times b+1\times a$. If we subtract both of the numbers, we get:
 $(10\times a+1\times b)-(10\times b+1\times a)$
= (10a + b) -  (10b + a)
= 10a  + b  -10b - a
= 9a - 9b
= 9 (a-b)
Observe that the result is a multiple of 9 and hence is exactly divisible by 9.
Similarly, for 3 digit numbers, the difference of the number and number obtained by reversing the digits (if abc is the number) will be given by:
$(100\times a+10\times b+c)-(100\times c+10\times b+a)$
=100a + 10b + c - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99 (a-c)
= $9\times 11(a-c)$ 
which is also divisible by 9.
Note that this happens with any 2 or 3 number.