Given a positive integer $k$, we need to find the minimum value of $N$ such that there exists a set $S$ of $2k + 1$ distinct positive integers satisfying the following conditions:
1. The sum of all elements in $S$ is greater than $N$.
2. Every subset of $S$ of size $k$ has a sum that is at most $\frac{N}{2}$.

To solve this problem, we construct the set $S$ using the $2k+1$ smallest positive integers, starting with 1, 2, 3, ..., up to $2k+1$.

### Step-by-Step Construction and Verification

1. Construct the Set:
Let $S = \{1, 2, 3, \ldots, 2k+1\}$.

2. **Calculate the Total Sum of $S$:**
The sum of the first $n$ integers is given by the formula:
$$\text{Sum}(S) = \frac{(2k+1)(2k+2)}{2}.$$

3. **Calculate $N$:**
Our goal is to ensure:
$$\text{Sum}(S) > N.$$

4. Subset Condition:
Every subset of size $k$ must have a sum that is at most $\frac{N}{2}$.

- The subset with the largest sum among subsets of size $k$ is obtained by taking the $k$ largest numbers from $S$, i.e., $\{k+2, k+3, \ldots, 2k+1\}$.

- The sum of this largest subset is:
$$\text{Max Subset Sum} = (k+2) + (k+3) + \ldots + (2k+1).$$
This is an arithmetic series whose sum is given by:
$$\text{Max Subset Sum} = \frac{k}{2}((k+2) + (2k+1)) = \frac{k}{2}(3k + 3).$$
Simplifying gives:
$$\text{Max Subset Sum} = \frac{3k^2 + 3k}{2}.$$

5. **Determine $N$:**
To satisfy the subset condition:
$$\frac{N}{2} \geq \frac{3k^2 + 3k}{2}.$$
Which implies:
$$N \geq 3k^2 + 3k.$$

6. Combine Results:
From the total sum condition:
$$\text{Sum}(S) > N.$$
Therefore, using:
$$\frac{(2k+1)(2k+2)}{2} > N \geq 3k^2 + 3k,$$
it suffices to have:
$$N = 2k^3 + 3k^2 + 3k.$$

Thus, the minimum value of $N$ is:
$$\boxed{2k^3 + 3k^2 + 3k}.$$