Find all prime numbers $p$ such that

$(x + y)^{19} - x^{19} - y^{19}$

is a multiple of $p$ for any positive integers $x$, $y$.
1. **Identify the problem**: We need to find all prime numbers \( p \) such that \((x + y)^{19} - x^{19} - y^{19}\) is a multiple of \( p \) for any positive integers \( x \) and \( y \).

2. **Use Fermat's Little Theorem**: For a prime \( p \), Fermat's Little Theorem states that \( a^{p-1} \equiv 1 \pmod{p} \) for any integer \( a \) not divisible by \( p \). This implies \( a^p \equiv a \pmod{p} \).

3. **Check specific primes**:
    - For \( p = 2 \):
      \[
      z^{19} \equiv z \pmod{2}
      \]
      This is true because any integer \( z \) modulo 2 is either 0 or 1, and \( 0^{19} \equiv 0 \) and \( 1^{19} \equiv 1 \).

    - For \( p = 3 \):
      \[
      z^{19} \equiv z \pmod{3}
      \]
      This is true because \( 19 \equiv 1 \pmod{2} \), and by Fermat's Little Theorem, \( z^3 \equiv z \pmod{3} \).

    - For \( p = 7 \):
      \[
      z^{19} \equiv z \pmod{7}
      \]
      This is true because \( 19 \equiv 5 \pmod{6} \), and by Fermat's Little Theorem, \( z^7 \equiv z \pmod{7} \).

    - For \( p = 19 \):
      \[
      z^{19} \equiv z \pmod{19}
      \]
      This is true because \( 19 \equiv 0 \pmod{18} \), and by Fermat's Little Theorem, \( z^{19} \equiv z \pmod{19} \).

4. **Check other primes**:
    - Consider the expression \( 2^{19} - 2 \):
      \[
      2^{19} - 2 = 524286 = 2 \cdot 3^3 \cdot 7 \cdot 19 \cdot 73
      \]
      This shows that the prime factors of \( 2^{19} - 2 \) are 2, 3, 7, 19, and 73.

    - Check \( p = 73 \):
      \[
      3^{19} \equiv 70 \pmod{73} \quad \text{and} \quad 2^{19} \equiv 2 \pmod{73}
      \]
      For \( x = 2 \) and \( y = 1 \):
      \[
      (x + y)^{19} - x^{19} - y^{19} \equiv 3^{19} - 2^{19} - 1^{19} \equiv 70 - 2 - 1 \equiv 67 \pmod{73}
      \]
      Since 67 is not congruent to 0 modulo 73, \( p = 73 \) does not work.

5. **Conclusion**: The only primes that work are 2, 3, 7, and 19.

The final answer is \(\boxed{2, 3, 7, 19}\).