Let's suppose there is a 2 digit number with both digits not the same and we reverse its digits order and then subtract from the original number and taking the absolute value of the difference obtained. Then is there a way to calculate for which number this difference is minimum.
And can there be a generalization for 3-digits numbers, 4-digits numbers and so on...?
The current answer does not answer OP's newly edited question, so I have decided to post a new answer.
Any 2-digit number can be expressed in the form $10a+b$ where $0\leq a, b<10$. Our final answer will be $(10a+b)-(10b+a)=9(a-b)$. Since $a\neq b$, $|9(a-b)|\geq9$. Any number satisfying $|a-b|=1$ works, like $10, 21, 34, $ etc.
If the question is generalised to a 3-digit number, any 3-digit number can be expressed in the form $100a+10b+c$ where $0\leq a, b, c<10$. Our final answer will be $(100a+10b+c)-(100c+10b+a)=99(a-c)$. Since, once again, $a\neq c$, $|99(a-c)|\geq99$. Any number satisfying $|a-c|=1$ works, like $130, 140, 768, $ etc.