A quadruplet of distinct positive integers $(a, b, c, d)$ is called $k$-good if the following conditions hold:

1. Among $a, b, c, d$, no three form an arithmetic progression.

2. Among $a+b, a+c, a+d, b+c, b+d, c+d$, there are $k$ of them, forming an arithmetic progression.

$a)$ Find a $4$-good quadruplet.
$b)$ What is the maximal $k$, such that there is a $k$-good quadruplet?
To solve this problem, we need to find a quadruplet of distinct positive integers \((a, b, c, d)\) that satisfies the given conditions. Let's break down the problem into two parts:

### Part (a)
We need to find a 4-good quadruplet. This means we need to find four distinct positive integers such that:
1. No three of them form an arithmetic progression.
2. Exactly 4 of the sums \(a+b, a+c, a+d, b+c, b+d, c+d\) form an arithmetic progression.

Let's consider the quadruplet \((1, 2, 4, 5)\).

1. **Checking for no three numbers forming an arithmetic progression:**
   - The possible triplets are \((1, 2, 4)\), \((1, 2, 5)\), \((1, 4, 5)\), and \((2, 4, 5)\).
   - None of these triplets form an arithmetic progression:
     - For \((1, 2, 4)\): \(2 - 1 \neq 4 - 2\)
     - For \((1, 2, 5)\): \(2 - 1 \neq 5 - 2\)
     - For \((1, 4, 5)\): \(4 - 1 \neq 5 - 4\)
     - For \((2, 4, 5)\): \(4 - 2 \neq 5 - 4\)

2. **Checking for sums forming an arithmetic progression:**
   - The sums are \(1+2=3\), \(1+4=5\), \(1+5=6\), \(2+4=6\), \(2+5=7\), \(4+5=9\).
   - We need to find 4 of these sums that form an arithmetic progression.
   - Consider the sums \(3, 5, 6, 7\):
     - The differences are \(5-3=2\), \(6-5=1\), \(7-6=1\).
     - This does not form an arithmetic progression.
   - Consider the sums \(3, 5, 7, 9\):
     - The differences are \(5-3=2\), \(7-5=2\), \(9-7=2\).
     - This forms an arithmetic progression with a common difference of 2.

Thus, the quadruplet \((1, 2, 4, 5)\) is a 4-good quadruplet.

### Part (b)
We need to determine the maximal \(k\) such that there exists a \(k\)-good quadruplet.

1. **Assume \(k \geq 5\):**
   - If \(k \geq 5\), then at least 5 of the sums \(a+b, a+c, a+d, b+c, b+d, c+d\) must form an arithmetic progression.
   - Consider the sums in increasing order: \(a+b < a+c < a+d < b+c < b+d < c+d\).
   - If 5 of these sums form an arithmetic progression, then the first two sums and the last two sums must be part of the arithmetic progression.

2. **Analyzing the sums:**
   - Let \(a+b\) and \(a+c\) be the first two elements of the arithmetic progression.
   - The third element could be \(b+c\) or \(a+d\), but this leads to contradictions:
     - If \(b+c\) is the third element, then \(a+c = 2b\), which is not possible since \(a, b, c, d\) are distinct.
     - If \(a+d\) is the third element, then \(b+d = 2c\), which is also not possible.

3. **Conclusion:**
   - Therefore, it is not possible to have 5 sums forming an arithmetic progression without violating the distinctness or the non-arithmetic progression condition among \(a, b, c, d\).
   - Hence, the maximal \(k\) is 4.

The final answer is \( \boxed{ k = 4 } \).