The answer is that n works iff the exponent of 2 in n is either 0 or odd (i.e., n is odd or n is of the form 2^odd odd).
The sum of the n consecutive positive integers starting at k is
n(n + 2k − 1)/2 = n²/2 + kn − n/2
The first form is more useful. Since we could start the consecutive integers at any integer, even at negative numbers, we can let n + 2k − 1 = x simply be any integer of opposite parity to n; then the sum is just nx/2.
This makes it clear when the conjecture fails, because if we let n be a multiple of 4 but not 8 (or 16 but not 32, etc.; it has a positive, even number of powers of 2), then x must be odd, and we are left with an odd number of factors of 2 in the sum, so it is not square.
Otherwise, it works. If n is odd, we let x = 2n. If n = a2b for odd a and b, then we let x = a. So it works for all odd n and for n divisible by 2 an odd number of times.