$$\boxed{309,\ 311}$$
Let $a$, $b$ and $c$ be each digit of $n$ in the hundreds, tens, and ones place respectively, then $n$ is described as $n = 100a + 10b + c$. Both $n + 2024$ and $n - 34$ have a prime number in the ones place, hence $c$ is $1$ or $9$.

In the case of $c = 1$, both integers
$$n + 2024 = 100(a + 20) + 10(b + 2) + 5, \quad n - 34 = 100a + 10(b - 4) + 7$$
have a prime number in tens place, hence $b = 1$. Similarly, both integers
$$n + 2024 = 100(a + 20) + 35, \quad n - 34 = 100(a - 1) + 77$$
have a prime number in the hundreds place, hence $a = 3$. Therefore, $n = 311$, and this $n$ satisfies the condition since $n + 2024 = 2335$ and $n - 34 = 277$.

In the case of $c = 9$, both integers
$$n + 2024 = 100(a + 20) + 10(b + 3) + 3, \quad n - 34 = 100a + 10(b - 3) + 5$$
have a prime number in the tens place, hence $b = 0$. Similarly, both integers
$$n + 2024 = 100(a + 20) + 33, \quad n - 34 = 100(a - 1) + 75$$
have a prime number in the hundreds place, hence $a = 3$. Therefore $n = 309$, and this $n$ satisfies the condition since $n + 2024 = 2333$ and $n - 34 = 275$.

We have proved that the two integers $n$ satisfying the condition are $309$, $311$.