Twenty pairs of integers are formed using each of the integers 1, 2, 3, ..., 40 once. The positive difference between the integers in each pair is 1 or 3. (For example, 5 can be paired with 2, 4, 6 or 8.) If the resulting differences are added together, the greatest possible sum is
Here is my attempt:
14, 25, 36, 710, 811, 912, 1316, 1417, 1518, 1922, 2023, 2124, 2528, 2629, 2730, 3134, 3235, 3336, 3740, 3839
The 3-digit numbers should be read as: 7 and 10 or 8 and 11.....etc. The 4-digit numbers as 13 and 16 or 14 and 17...etc.
There is an "absolute" difference of 3 between all the pairs, except the last one. Therefore, the maximum sum I get is:
19 x 3 + 1 =58
+118608 I do not know if I understand the question but if the biggest difference between the pairs is 3 then the pairs could be
1,4
2,5
3,6
7,10
8,11
9,12
13,16
14,17
15,18
19,22
20,23
21,24
25,28
26,29
27,30
31,34
32,35
33,36
37,40
38,39
19*3+1 = 58
So I get the same 'guess' answer as guest does. 
