Let $\{x_1, x_2, x_3, ..., x_n\}$ be a set of $n$ distinct positive integers, such that the sum of any $3$ of them is a prime number. What is the maximum value of $n$?
1. **Assume $n \geq 5$ and derive a contradiction:**
   - Let $\{x_1, x_2, x_3, \ldots, x_n\}$ be a set of $n$ distinct positive integers such that the sum of any three of them is a prime number.
   - Consider the residues of these integers modulo 3. There are three possible residues: 0, 1, and 2.

2. **Pigeonhole Principle:**
   - By the Pigeonhole Principle, if $n \geq 5$, then at least two of these integers must have the same residue modulo 3.
   - Suppose we have at least two integers with the same residue modulo 3. Without loss of generality, assume $x_1 \equiv x_2 \pmod{3}$.

3. **Sum of three integers with the same residue:**
   - Consider the sum of three integers with the same residue modulo 3. For example, if $x_1 \equiv x_2 \equiv x_3 \pmod{3}$, then:
     \[
     x_1 + x_2 + x_3 \equiv 0 + 0 + 0 \equiv 0 \pmod{3}
     \]
   - Since the sum of three integers with the same residue modulo 3 is divisible by 3, it cannot be a prime number unless the sum is 3 itself. However, since $x_1, x_2, x_3$ are distinct positive integers, their sum must be greater than 3.

4. **Contradiction:**
   - Therefore, if $n \geq 5$, it is impossible for the sum of any three integers to be a prime number, leading to a contradiction.

5. **Verification for $n = 4$:**
   - Consider the set $\{1, 3, 7, 9\}$:
     - Sum of any three elements:
       \[
       1 + 3 + 7 = 11 \quad (\text{prime})
       \]
       \[
       1 + 3 + 9 = 13 \quad (\text{prime})
       \]
       \[
       1 + 7 + 9 = 17 \quad (\text{prime})
       \]
       \[
       3 + 7 + 9 = 19 \quad (\text{prime})
       \]
   - All sums are prime numbers, so $n = 4$ is possible.

6. **Conclusion:**
   - The maximum value of $n$ such that the sum of any three elements is a prime number is 4.

The final answer is $\boxed{4}$.