Problem: Let $k > 0$ be a fixed integer.
Compute the minimum integer $N$ (in terms of $k$) for which
there exists a set of $2k+1$ distinct positive integers
that has sum greater than $N$,
but for which every subset of size $k$ has sum at most $N/2$.

Solution: The answer is $N = k(2k^2+3k+3)$ given by
\[ S = \left\{ k^2+1, k^2+2, \dots, k^2+2k+1 \right\}. \]

To show this is best possible,
let the set be $S = \{ a_0 <  a_1 < \dots < a_{2k} \}$
so that the hypothesis becomes
\begin{align*}
  N + 1 &\le a_0 + a_1 + \dots + a_{2k} \\
  N/2 &\ge a_{k+1} + \dots + a_{2k}.
\end{align*}
Subtracting twice the latter from the former gives
\begin{align*}
  a_0 &\ge 1 + (a_{k+1}-a_1) + (a_{k+2}-a_2) + \dots
  + (a_{2k} - a_k) \\
  &\ge 1 + \underbrace{k + k + \dots + k}_{k \text{ terms}} = 1 + k^2.
\end{align*}
Now, we have
\begin{align*}
  N/2 &\ge a_{k+1} + \dots + a_{2k} \\
  &\ge (a_0 + (k+1)) + (a_0 + (k+2)) + \dots + (a_0 + 2k) \\
  &= k \cdot a_0 + \left( (k+1) + \dots + 2k \right) \\
  &\ge k(k^2+1) + k \cdot \frac{3k+1}{2}
\end{align*}
so $N \ge k(2k^2+3k+3)$.

\begin{remark*}
  The exact value of $N$ is therefore very superficial.
  From playing with these concrete examples
  we find out we are essentially
  just trying to find an increasing set $S$ obeying
  \[ a_0 + a_1 + \dots + a_k > a_{k+1} + \dots + a_{2k} \qquad (\star) \]
  and indeed given a sequence satisfying these properties
  one simply sets $N = 2(a_{k+1} + \dots + a_{2k})$.
  Therefore we can focus almost entirely on $a_i$ and not $N$.
\end{remark*}

\begin{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
  \begin{itemize}
    \ii $\{2,3,4\}$ is optimal for $k=1$
    \ii $\{5,6,7,8,9\}$ is optimal for $k=2$,
    \ii $\{10,11,12,13,14,15,16\}$ is optimal for $k=3$.
  \end{itemize}
  In all the examples,
  the $a_i$ are an arithmetic progression of difference $1$,
  so that $a_j - a_i \ge j-i$ 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.
\end{remark*}