, 2012

In an ordered list of five integers, the median is the number in the middle or third position.

Thus if we let the set of integers be $a,b,c,d,e$, ordered from smallest to largest, then $c=18$.

Since the average is fixed (at 20), $e$ (the largest number in the set) is largest when $a,b$ and $d$ are as small as possible.

Since the numbers in the set are different positive integers, the smallest that $a$ can be is 1 and the smallest that $b$ can be is 2.

Our set of integers is now $1,2,18,d,e$.

Again, we want $d$ to be as small as possible, but it must be larger than the median 18.

Therefore, $d=19$.

Since the average of the 5 integers is 20, then the sum of the five integers is $5\times20=100$.

Thus, $1+2+18+19+e=100$ or $40+e=100$, and so $e=60$.
The greatest possible integer in the set is 60.