Take the same basic idea as in Matchstick Frenzy II - A heap of a positive integer number of matches (that is, no broken matches) is divided into n groups.

This time, we take from the first group the square of the number of matches in the second group and add to the second, then take from the second group the square of the number of matches in the third group and add to the third, and so on up to the nth group.

Finally we take from the nth group the square of the number of matches left in the first group and add them to the first group to make the number of matches in each group the same positive integer.

Now for the question: how many matches were there in the first group to start with?

The closest to a solution I found for more than two groups is for three groups. The first group starts with 299 matches, the second group with 17 matches, and the third (nth) group starts with 14 matches. With a final total of 110 matches in each group.

Group 1:
  299
- 289 = 17² <-take from 1st group square of 2nd group
=====
   10

Group 2:
   17
+ 289 = 17² <-add to the 2nd group
=====
  306
- 196 = 14² <-take from 2nd group square of 3nd group
=====
  110

Group 3 (N):
   14
+ 196 = 14² <-add to the 3nd group
=====
  210
- 100 = 10² <-take from 3rd group square of 1st group
=====
  110

Group 1:
   10
+ 100 = 10² <-add to the 1st group
=====
  110 <--- Each group has the same number of matches