Japan Mathematical Olympiad

$$\boxed{309,311}$$ 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 $$n+2024=100(a+20)+10(b+2)+5,n-34=100a+10(b-4)+7$$ have a prime number in tens place, hence $b=1$. Similarly, both integers $$n+2024=100(a+20)+35,n-34=100(a-1)+77$$ 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 $$n+2024=100(a+20)+10(b+3)+3,n-34=100a+10(b-3)+5$$ have a prime number in the tens place, hence $b=0$. Similarly, both integers $$n+2024=100(a+20)+33,n-34=100(a-1)+75$$ 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 $309$, $311$.