The maximum possible difference between two 2-digit positive integers is obtained when one integer is in 90s and other is in 10s i.e one of the integer has 9 at ten's place and other has 1 at ten's place.

Let x be the unit's digit of 1 integer and y be the unit's digit of other.

So, our numbers are 90+x and 10+y

Let m be the difference between these two integers.

Then,
\(90+x - (10+y)=m\)
or \(80+x-y=m\)
or \(x-y=m-80\) .......(1)

Now, the integers have their respective ten's digits exchanged. So, the new numbers are 10+x and 90+y.

It is given that the difference changes by 4. To maximize the difference of original numbers, the new difference should reduce by 4. So, our new difference is m-4

\(90+y-(10-x)=m-4\)
or \(80+y-x=m-4\)
or \(y-x=m-84\)
or \(x-y=84-m\) ........(2)

From 1 and 2 we have,
\(m-80=84-m\)
or \(2m=164\)
or \(m=82\)

Answer: C