The minimum is $N = 2k^3 + 3k^2 + 3k$. The set
$$\{k^2+1, k^2+2, \dots, k^2+2k+1\}$$
has sum $2k^3 + 3k^2 + 3k + 1 = N + 1$ which exceeds $N$, but the sum of the $k$ largest elements is only $(2k^3 + 3k^2 + 3k)/2 = N/2$. Thus this $N$ is such a value.

Suppose $N < 2k^3 + 3k^2 + 3k$ and there are positive integers $a_1 < a_2 < \dots < a_{2k+1}$ with $a_1 + a_2 + \dots + a_{2k+1} > N$ and $a_{k+2} + \dots + a_{2k+1} \le N/2$. Then
$$(a_{k+1} + 1) + (a_{k+1} + 2) + \dots + (a_{k+1} + k) \le a_{k+2} + \dots + a_{2k+1} \le N/2 < \frac{2k^3 + 3k^2 + 3k}{2}.$$
This rearranges to give $2k a_{k+1} \le N - k^2 - k$ and $a_{k+1} < k^2 + k + 1$. Hence $a_{k+1} \le k^2 + k$. Combining these we get
$$2(k+1)a_{k+1} \le N + k^2 + k.$$
We also have
$$(a_{k+1} - k) + \dots + (a_{k+1} - 1) + a_{k+1} \ge a_1 + \dots + a_{k+1} > N/2$$
or $2(k+1)a_{k+1} > N + k^2 + k$. This contradicts the previous inequality, hence no such set exists for $N < 2k^3 + 3k^2 + 3k$ and the stated value is the minimum.