Given a positive integer n, what is the largest k such that
the numbers 1,2,...,n can be put into k boxes so that
the sum of the numbers in each box is the same?

[When
n = 8, the example {1,2,3,6},{4,8},{5,7} shows that
the largest k is at least 3.]

for n odd, k=(n+1)/2

for n even, k=n/2

Attempted (rather clumsy) proof:

Let's say s is the sum in each box.

The smaller s is, the larger k is, since

(s k) = n (n+1)/2 = sum({ i= 1 to n} i )

The smallest possible sum is n and the next smallest possible sum is n+1, since n must itself be in some box.

For the odd n, s can be n, with n sitting in its own box, and the other boxes must hold ( n-i, i ), and so k = (n+1)/2, a maximum.

For the even n, n will not work as the sum s, because then we are again forced to make boxes of the form ( n-i, i) and this leaves n/2 remaining with no partner. However s = n+1 works with n/2 pairs of the form:

( n+1-i, i ), and k = n/2.

QED

*Edited on ***October 1, 2018, 8:37 am**