Let $k$ be a given positive integer, find the smallest positive integer $N$, such that there exists a set composed of $2k+1$ different positive integers, the sum of whose elements is greater than $N$, but the sum of the elements of any $k$-element subset is at most $\frac{N}{2}$.
Let $N$ be a positive integer that satisfies the condition, and let $\left\{a_{1}, a_{2}, \cdots, a_{2 k+1}\right\}$ be a $(2 k+1)$-element set that meets the problem's requirements, where $a_{1}N$.

Notice that $a_{1}, \cdots, a_{2 k+1} \in \mathbf{N}^{*}$, so $a_{k+1+i} \geqslant a_{k+1}+i, a_{k+1-i} \leqslant a_{k+1}-i$, where $i=1,2, \cdots, k$. Therefore,
$$
\left\{\begin{array}{l}
\left(a_{k+1}+1\right)+\left(a_{k+1}+2\right)+\cdots+\left(a_{k+1}+k\right) \leqslant \frac{N}{2} \\
\left(a_{k+1}-k\right)+\left(a_{k+1}-(k-1)\right)+\cdots \\
\quad+\left(a_{k+1}-1\right)+a_{k+1}>\frac{N}{2} .
\end{array}\right.
$$

Here, we use $a_{1}+\cdots+a_{k+1}=\left(a_{1}+\cdots+a_{2 k+1}\right)$
$$
-\left(a_{k+2}+\cdots+a_{2 k+1}\right)>N-\frac{N}{2}=\frac{N}{2} .
$$

From (1) and (2), we have $\frac{N+k^{2}+k}{2(k+1)}<a_{k+1} \leqslant \frac{N-k^{2}-k}{2 k}$. Since $a_{k+1} \in \mathbf{N}^{*}$, the interval $\left(\frac{N+k^{2}+k}{2(k+1)}, \frac{N-k^{2}-k}{2 k}\right]$ contains a positive integer. If $N$ is odd, then $\frac{N-k^{2}-k-1}{2 k} \geqslant \frac{N+k^{2}+k+1}{2(k+1)}$; if $N$ is even, then $\frac{N-k^{2}-k}{2 k} \geqslant \frac{N+k^{2}+k+2}{2(k+1)}$. Calculating each case, we find that $N \geqslant 2 k^{3}+3 k^{2}+3 k$.
Additionally, for $N=2 k^{3}+3 k^{2}+3 k$, the set
$$
\left\{k^{2}+1, k^{2}+2, \cdots, k^{2}+2 k+1\right\}
$$

has a sum of all its elements $=2 k^{3}+3 k^{2}+3 k+1=N+1$, and the sum of the largest $k$ elements $=k^{3}+(k+2)+\cdots+(2 k+1)=k^{3}+\frac{(3 k+3) k}{2}=\frac{N}{2}$.
Therefore, the minimum value of $N$ is $2 k^{3}+3 k^{2}+3 k$.