1. Let the largest two elements in the set be $a$ and $b$ such that $a < b$. For all other elements in the set, let $x$ be an element such that $x < a$.

2. The sum of all other elements in the set is at most $\frac{(a-1)a}{2}$. This is because the sum of the first $a-1$ positive integers is $\frac{(a-1)a}{2}$.

3. According to the problem, $\frac{8}{19}$ of the product of the largest two elements is not greater than the sum of the other elements. Therefore, we have:
$$\frac{8}{19} \cdot a \cdot b \leq \frac{(a-1)a}{2}$$

4. Simplify the inequality:
$$\frac{8ab}{19} \leq \frac{a(a-1)}{2}$$

5. Multiply both sides by 38 to clear the fractions:
$$16ab \leq 19a(a-1)$$

6. Divide both sides by $a$ (assuming $a \neq 0$):
$$16b \leq 19(a-1)$$

7. Solve for $b$:
$$b \leq \frac{19(a-1)}{16}$$

8. Since $b$ must be greater than $a$, we have:
$$a < b \leq \frac{19(a-1)}{16}$$

9. To find the minimum possible value of $a$, we need to find the smallest integer $a$ such that:
$$a < \frac{19(a-1)}{16}$$

10. Simplify the inequality:
$$16a < 19(a-1)$$
$$16a < 19a - 19$$
$$0 < 3a - 19$$
$$19 < 3a$$
$$a > \frac{19}{3}$$
$$a > 6.33$$

11. Since $a$ must be an integer, the smallest possible value for $a$ is 7. However, we need to check if this value satisfies the original inequality.

12. If $a = 7$:
$$b \leq \frac{19(7-1)}{16} = \frac{114}{16} = 7.125$$
Since $b$ must be an integer and greater than $a$, $a = 7$ does not work.

13. Try $a = 8$:
$$b \leq \frac{19(8-1)}{16} = \frac{133}{16} = 8.3125$$
Since $b$ must be an integer and greater than $a$, $a = 8$ does not work.

14. Try $a = 12$:
$$b \leq \frac{19(12-1)}{16} = \frac{209}{16} = 13.0625$$
Since $b$ must be an integer and greater than $a$, the smallest possible value for $b$ is 13.

Therefore, the minimum possible value of the largest number in the set is $13$.

The final answer is $\boxed{13}$.