I found two solutions.

The 5 numbers, followed by the 10 sums:

(20, 23, 25, 27, 31)
 [43, 45, 47, 48, 50, 51, 52, 54, 56, 58]

(21, 23, 24, 27, 31)
 [44, 45, 47, 48, 50, 51, 52, 54, 55, 58]

The fact that the middle 6 sums show no duplicates implies, and I'm fairly certain proves, that the set of 5 integers also has no duplicates, but for completeness I decided to search with the possibility of duplicates in mind.

If we number the 10 sums in order from 1 to 10, and call the set of 5 integers a,b,c,d,e from smallest to largest, then:

if a+d=b+c then 3,4 are same
if a+e=b+c then 4,5 are same
if a+e=b+d then 5,6 are same
if a+e=c+d then 6,7 are same
if b+e=c+d then 7,8 are same

I didn't do it, but I suspect that requiring all 5 equations to be false will prove that none of a,b,c,d,e are the same.

--- code follows ---

from itertools import combinations
from itertools import combinations_with_replacement

def testSums(aList):
    """ input is a list of integers
    returns a list of sum of any 2 of these integers,
    sorted in order from smallest to largest """
    ans = []
    for group_of_2 in combinations(aList,2):
        ans.append(sum(group_of_2))
    return sorted(ans)

match = [47, 48, 50, 51, 52, 54]

myList = [i for i in range(1,60)]

for group_of_5 in combinations_with_replacement(myList,5):
    middle6 = testSums(group_of_5)[2:8]
    if middle6 == match:
        print(group_of_5,'\n', testSums(group_of_5))