If each pile adds up to k, we need 2k= 1+… +n = n(n+1)/2, so n(n+1) =4k.
But since in 2 consecutive numbers, one is odd and one is even, it’s necessary that 4 divides n or 4 divides n+1.
We show that if 4 divides n, it is possible:
If n=4k: in 1, 2, 3 ….. 4k-2,4k-1,4k
The Gauss trick shows that adding 1+ 4k, 2+ (4k-1), until k+(3k+1) gives k(4k+1)
in the other set adding up : (k+1) + 3k,…, (k+k) + 2k+1 gives k(4k+1), the same
Now we show that if 4 divides n+1, that is, n=4k+3, it is also possible
The first set must have 4k+3, 1 and 4k+2,2 and 4k+1 until k, 3k+3: that adds up to (k+1)(4k+3)
The other set has k+1 up to 3k+2: this adds up to (k+1)(4k+3) as well.
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