1. Understanding the Problem:
We need to define a function $f(i)$ for any positive integer $i$. The function $f(i)$ is the smallest positive integer $j$ such that there exist pairwise distinct positive integers $a, b, c,$ and $d$ with the greatest common divisors (gcd) of each pair being pairwise distinct and equal to $i, i+1, i+2, i+3, i+4,$ and $j$ in some order. If no such $j$ exists, then $f(i) = 0$.

2. Key Observation:
For any integers $a, b, c,$ and $d$, the gcds of the pairs must satisfy the condition that they are pairwise distinct and equal to $i, i+1, i+2, i+3, i+4,$ and $j$. This implies that the gcds must be distinct and cover a range of six consecutive integers.

3. Using Modulo 12:
We observe that the gcds must be distinct and cover a range of six consecutive integers. This suggests that the function $f(i)$ might have a periodic behavior modulo 12. We will compute $f(i)$ for $i$ from 1 to 12 and then use this periodicity to find the sum $f(1) + f(2) + \dots + f(2019)$.

4. **Computing $f(i)$ for $i$ from 1 to 12:**
- $f(1) = 14$
- $f(2) = 9$
- $f(3) = 0$
- $f(4) = 0$
- $f(5) = 0$
- $f(6) = 3$
- $f(7) = 2$
- $f(8) = 0$
- $f(9) = 0$
- $f(10) = 1$
- $f(11) = 0$
- $f(12) = 0$

5. Periodic Behavior:
From the above computations, we observe that $f(i+12) = f(i)$ for $i > 2$. This periodicity allows us to extend the values of $f(i)$ for $i$ up to 2019.

6. **Summing $f(i)$ from 1 to 2019:**
- The sequence $f(i)$ repeats every 12 numbers.
- We need to sum $f(i)$ for 2019 terms.
- There are $\left\lfloor \frac{2019}{12} \right\rfloor = 168$ complete cycles of 12 terms.
- The sum of one cycle (from $f(1)$ to $f(12)$) is $14 + 9 + 0 + 0 + 0 + 3 + 2 + 0 + 0 + 1 + 0 + 0 = 29$.
- The total sum for 168 cycles is $168 \times 29 = 4872$.

7. Handling the Remaining Terms:
- There are $2019 \mod 12 = 3$ remaining terms.
- These terms are $f(1) + f(2) + f(3) = 14 + 9 + 0 = 23$.

8. Final Sum:
- The total sum is $4872 + 23 = 4895$.

The final answer is $\boxed{4895}$