To solve the problem, we need to find all 5-tuples of different four-digit integers with the same initial digit such that the sum of the five numbers is divisible by four of them. Let's denote the five numbers as $a, b, c, d, e$ where $a, b, c, d, e$ are four-digit integers and $a > b > c > d > e$.

1. Initial Setup:
We need to find $a, b, c, d, e$ such that:
$$a + b + c + d + e \text{ is divisible by } a, b, c, d \text{ or } e$$

2. Divisibility Conditions:
We can write the conditions as:
$$a + b + c + d + e = ka \quad \text{for some integer } k$$
$$a + b + c + d + e = lb \quad \text{for some integer } l$$
$$a + b + c + d + e = mc \quad \text{for some integer } m$$
$$a + b + c + d + e = nd \quad \text{for some integer } n$$
$$a + b + c + d + e = pe \quad \text{for some integer } p$$

3. Handling Equality Cases:
We will handle the equality cases one by one:

- **Case 1: $a = b = c = d = e$**
$$a + b + c + d + e = 5a$$
This case is not possible since $a, b, c, d, e$ are distinct.

- **Case 2: $a = b = c = d \neq e$**
$$4a + e = ka$$
Testing values for $k = 2, 3, 4, 5, 6$, none of these work. $k > 6$ is impossible by size arguments.

- **Case 3: $a = b = c \neq d \neq e$**
$$3a + d + e = ka$$
Since $a > b > c > d > e$, $k > 3$ by size arguments and $k > 4$ is impossible. So $k = 3$ implying $2a = d + e$. This contradicts $a > b > c > d > e$.

- **Case 4: $a = b \neq c \neq d \neq e$**
$$2a + c + d + e = ka$$
Since $a > b > c > d > e$, $k = 2, 3$ and no solutions exist.

- **Case 5: $a = b \neq c = d \neq e$**
$$2a + 2c + e = ka$$
Testing the cases, none of these work.

4. Strict Inequality Case:
Since we have discarded the equality cases, we now consider $a > b > c > d > e$. We need to solve:
$$ka = a + b + c + d + e < 4a$$
So $k = 2, 3$ for which no solutions exist other than:
$$(a, b, c, d, e) = (60t, 70t, 84t, 101t, 105t), \; t \in \{17, 18, 19\}$$
Hence, the only solutions are:
$$\boxed{(a, b, c, d, e) = (60t, 70t, 84t, 101t, 105t), \; t \in \{17, 18, 19\} \text{ and its permutations}}$$