Answer: All odd $1\leq k \leq 12$ except $11$. Let $S$ be the set with $12$ non-zero real numbers. If $k$ is even, then the product must be positive and the sum must be negative in the set $\{-1,-1,...,-1\}$ which gives contradiction. Claim: $k=1,3,5$ is achievable. Assume that there exists a set where at least one of $k=1,3,5$ is not achievable. If there exists $k$ negative numbers, then that subset has negative product and sum which is not possible because of our assumption. If there exists $k$ positive numbers, then that subset has positive product and sum which is not possible because of our assumption, again. So $12\leq 2k\iff 6\leq k$ which gives a contradiction. For $k=7$ Assume that there exists a set where $k=7$ is not achievable. There must be at most $6$ positive and $6$ negative real numbers. Since there are $12$ numbers, there are exactly $6$ positive and $6$ negative numbers. If we take a set $T$ which contains $4$ negative and $3$ positive numbers, since the product is positive, the sum must be negative. By changing a positive number in $T$ with a negative number which is not in $T$, we have negative sum and negative product which gives a contradiction. For $k=9$ Assume that there exists a set where $k=9$ is not achievable. There must be at most $8$ positive, $8$ negative real numbers. If there exists exactly $4$ positive and $8$ negative real numbers, then we can choose $4$ positive and $5$ negative numbers. The product is negative so the sum must be positive. After changing a positive in the subset with a negative which is not in the subset, the product becomes positive and the sum stays positive which gives a contradiction. If there are at least $5$ positive numbers, then we take a subset which contains $4$ positive and $5$ negative numbers. After changing a negative in the subset with a positive which is not in the subset, the product becomes positive and the sum stays positive which contradicts. For $k=11$ The set $(-19,-19,2,2,2,2,2,2,2,2,2,2)$ satisfies. Thanks, Below.

For k=11 @above how about the set (-19,-19, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2) Choose any 11 of them. You either get 1 or 2 -19s. If you have 1 the sum is 1 while the product is negative. If you have 2 the sum is negative but the sum is clearly positive. The other parts of the solution are correct