Solution 1 a) Notice that $\left\{2019!\cdot 1^{2019!}, 2019!\cdot 2^{2019!}, \ldots, 2019!\cdot 2019^{2019!}\right\}$ is such a set. Observe that if all the elements are divisible by 2019! then the arithmetic means will be integer for all the subsets. Also, if $A$ is a set such that the geometric means are integer for all non-empty subsets and the set $B$ is obtained from the set $A$ by multiplying each element with with a given integer $c$ then all the non-empty subsets of $B$ will have an integer geometric mean, since

$$\sqrt[k]{c a_{i_{1}} c a_{i_{2}} \cdots c a_{i_{k}}}=c \sqrt[k]{a_{i_{1}} a_{i_{2}} \cdots a_{i_{k}}}$$

It is thus sufficient to find a set of 2019 positive integers such that the geometric mean of every non-empty subset in an integer. Now, for an integer $a$ the number $\sqrt[k]{a^{2019!}}=a^{\frac{2019!}{k}}$ for all integers $1 \leq k \leq 2019$ so $\left\{1^{2019!}, 2^{2019!}, \ldots, 2019^{2019!}\right\}$ is a set such that the geometric mean of every non-empty subset is an integer.

b) Assume there exist such a set $A$ and let $n, m, a_{1}, a_{2}, \ldots, a_{m-1}$ be distinct elements in $A$ with $n<m$. Then $\frac{n+a_{1}+a_{2}+\cdots+a_{m-1}}{m}$ and $\frac{m+a_{1}+a_{2}+\cdots+a_{m-1}}{m}$ are integers and also their difference

$$\frac{m+a_{1}+a_{2}+\cdots+a_{m-1}}{m}-\frac{n+a_{1}+a_{2}+\cdots+a_{m-1}}{m}=\frac{m-n}{m}$$

Therefore, we have $\frac{n}{m}$ is an integer and since $m$ and $n$ are positive integers we have $m \leq n$ which is a contradiction.