In a magic triangle of addition, each of six natural numbers from 21 to 26 is placed in one of the six empty circles such that the sum $$S$$ of the three numbers from the circles along each side would be the same. Then the maximum possible value of $$S$$ is

Adding the three side sums counts every number once and each of the three corner numbers twice, so $$3S = (21 + 22 + \cdots + 26) + (\text{sum of corners}) = 141 + C$$. To make $$S$$ largest take the largest numbers at the corners, $$C = 24 + 25 + 26 = 75$$, giving $$3S = 216$$ and $$S = 72$$. This is achievable, since 23, 21 and 22 placed between the pairs $$(24, 25)$$, $$(25, 26)$$ and $$(26, 24)$$ make every side add to 72.