Let $k>0$ be a fixed integer. Compute the minimum integer $N$ (in terms of $k$ ) for which there exists a set of $2 k+1$ distinct positive integers that has sum greater than $N$, but for which every subset of size $k$ has sum at most $N / 2$.
The answer is $N=k\left(2 k^{2}+3 k+3\right)$ given by
$$
S=\left\{k^{2}+1, k^{2}+2, \ldots, k^{2}+2 k+1\right\} .
$$

To show this is best possible, let the set be $S=\left\{a_{0}a_{k+1}+\cdots+a_{2 k}
$$
and indeed given a sequence satisfying these properties one simply sets $N=2\left(a_{k+1}+\cdots+a_{2 k}\right)$. Therefore we can focus almost entirely on $a_{i}$ and not $N$.

Remark. It is relatively straightforward to figure out what is going on based on the small cases. For example, one can work out by hand that
- $\{2,3,4\}$ is optimal for $k=1$
- $\{5,6,7,8,9\}$ is optimal for $k=2$,
- $\{10,11,12,13,14,15,16\}$ is optimal for $k=3$.

In all the examples, the $a_{i}$ are an arithmetic progression of difference 1 , so that $a_{j}-a_{i} \geq j-i$ is a sharp for all $i<j$, and thus this estimate may be used freely without loss of sharpness; applying it in $(\star)$ gives a lower bound on $a_{0}$ which is then good enough to get a lower bound on $N$ matching the equality cases we found empirically.