To determine whether there exists a finite set $A$ of positive integers such that for each $a \in A$, at least one of the numbers $2a$ or $\frac{a}{3}$ belongs to $A$, we proceed as follows:

Assume for the sake of contradiction that such a finite set $A$ exists. We will focus on the properties of the elements within this set.

1. Property of Multiplication by 2:

If $a \in A$, then $2a$ must also be included in $A$. This means that starting from any element $a$, repeatedly multiplying by 2 gives additional elements that must also be in $A$. This forms an infinite sequence $a, 2a, 4a, 8a, \ldots$.

2. Property of Division by 3:

Similarly, if $a \in A$ and $a$ is divisible by 3, then $\frac{a}{3}$ must be in $A$. Continuously dividing by 3 (if possible) forms another sequence. However, dividing by 3 can only continue while the result remains a positive integer.

Given that $A$ is a finite set, eventually, these procedures of multiplying by 2 and dividing by 3 (when possible) must terminate.

3. Contradiction from Finite Assumption:

Let's explore the implication of having such operations in a supposed finite set $A$:

- Consider the largest element $b \in A$. Applying the doubling process from any element less than or equal to $b$ will generate elements $2b, 4b, \ldots$ potentially larger than $b$. Therefore, these elements must exist in $A$, forcing $A$ to expand beyond $b$, contradicting the finiteness of $A$.

- Suppose all elements $a \in A$ cannot be divided by 3. Then the requirement $\frac{a}{3} \in A$ can never be satisfied for any $a$. This requires $A$ to empty, further implying $A$ cannot exist.

Thus, given any initial assumption of the set $A$ being finite, we derive contradictions via the infinite generation of elements through multiplication or unfeasible satisfaction of conditions via division by 3. Therefore:

The answer is that no such finite set $A$ exists. Thus, the final conclusion is:

$$\boxed{\text{No}}$$