Is This Even Possible?

I claim that all multiples of $4$ work, and no others.

First, to see that multiples of $4$ work, let $n=4k$ . We consider two cases.

• Case 1.1: $n$ is a multiple of $8$ .

Then consider $2k, 2, 1, 1, \ldots, 1, -1, -1, \ldots, -1$ , with $k-2$ instances of $1$ and $3k$ instances of $-1$ . The sum is $2k + 2 + (k-2) \cdot 1 + (3k) \cdot (-1) = 0$ , and the product is $(2k)(2)(1)^{k-2}(-1)^{3k} = 4k$ because $3k$ is even.

• Case 1.2: $n$ is not a multiple of $8$ .

Then consider $2k, -2, 1, 1, \ldots, 1, -1, -1, \ldots, -1$ , with $k$ instances of $1$ and $3k-2$ instances of $-1$ . The sum is $2k - 2 + k \cdot 1 + (3k-2) \cdot (-1) = 0$ , and the product is $(2k)(-2)(1)^k(-1)^{3k-2} = 4k$ because $3k-2$ is odd.

Now, to prove that non-multiples of $4$ don't work, we consider two cases.

• Case 2.1: $n$ is odd.

Then all of the integers must be odd (because their product is an odd number $n$ ). But then the sum is also odd (since there are an odd number of odd integers), so it cannot be $0$ .

• Case 2.2: $n$ is even but not a multiple of $4$ .

Then exactly one of the integers must be even, and the rest odd (because there is exactly one instance of the factor $2$ ). But then the sum is odd (since there is $n-1$ odd integers, and $n-1$ is odd), so it cannot be $0$ .

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This shows that all multiples of $4$ work and no other. There are $\boxed{249}$ multiples of $4$ below $1000$ .