Enigma
This game, as well as Finding The Faulty Coin and The Chain, comes from an enigma which furnishes valuable mathematical information. We have 4 standard weights to measure all whole numbers from 1 to 40 (which is evidently the sum of these 4 weights). The information is precise and complete, but the enigma could have been more "enigmatic" by asking for the smallest number (as in The Chain) rather than by specifying the number of weights.

Specific To General
The smallest standard weight must evidently be "1",
         the next weight can be "3", instead of "2", because the "1" can be subtracted from the "3",
                   and all the smaller weights can be subtracted from the larger ones.
If the standard weights (in this case, 1, 3, 9, 27) are indicated W⁰, W¹, W², W³, where
         W⁰ is evidently "1",
         W¹ is the base of the multiplicative system (in this specific case "3"),
         and the others are the ensuing exponentials (in this specific case of "3"),
                   the sum of the weights up to Wⁿ will be (3^{(n + 1)} - 1) / (3 - 1),
                             from the general formula (W^{(n + 1)} - 1) / (W¹ - 1).
         If there is only 1 weight (n = 0), its weight will be 1, and the total will also be 1 ((3 - 1) / (3- 1)).
         If there are 2 weights (n = 1), their weight will be 1 and 3 for a total of 4 ((9 - 1) / (3- 1)).
         If there are 3 weights (n = 2), their weight will be 1, 3, and 9, for a total of 13 ((27 - 1) / (3- 1)).
         If there are 4 weights (n = 3), their weight will be 1, 3, 9, and 27, for a total of 40 ((81 - 1) / (3- 1)).
         If there are 5 weights (n = 4), their weight will be 1, 3, 9, 27, and 81, for a total of 121 ((243 - 1) / (3- 1)).

Problem-solving
This is a case of logic in the sense that there is only one unique solution. However, reasoning the answer is not as simple as it was in the case of Finding The Faulty Coin (with its 9 = 3²). We are obliged to build up the process from scratch as we did in the previous paragraph and see which solution gives a total of 40 (with 4 weights of 1, 3, 9, and 27). This might be considered trial-and-error by some but it remains a planified, orderly, if slightly fastidious process.