.
Solution by Benyoucef Sidoummou (Algeria). (Translated from the French.)
For the solution we use the formula for the sum of the first n counting numbers,
We shall denote the sum of the first n - 1 counting numbers by s; that is, s = n(n - 1)/2. Note that the nth group of numbers in our problem contains n numbers. One must find the first of them: it is 1 + (the last number of the preceding group). Moreover, the last number of the preceding group (that is, of the n - 1st group) is s because it is the sum of the 1 number in group 1, plus the two in group 2, plus the three in group 3, and so on up to the n - 1 numbers in group n - 1. It follows that the sum of the numbers in the nth group is
which equals ns + (the sum from 1 to n ) =
We received nice solutions also from Midhun Chandran, Tony Kao, Yueng Sze Man, Juan Mir Pieras (Spain), Alexander Potapenko (Russia), and Yoann. All were quite similar to our featured solution. We also received several partial solutions; the one from bantum_1_cwa looked as if he or she had the right idea, but we are looking for complete solutions.
We took the problem from Crux Mathematicorum 7 (1981), page 63. It was #518, proposed by Charles W. Trigg.