Answer: $n=240$.

Solution. In fact, it is required that the fourth powers of all prime numbers greater than 10 give the same remainder when divided by $n$. Indeed, if the fourth powers of some two numbers give different remainders, then one can take $n-1$ of the first number and one of the second, then the sum of the fourth powers of these numbers will not be divisible by $n$.

We will prove that the fourth power of any prime number $p$, greater than 10, when divided by 3, 5, and 16, gives a remainder of 1, then it will also give a remainder of 1 when divided by $3 \cdot 5 \cdot 16=240$. We will use the fact that $p^{4}-1=\left(p^{2}-1\right)\left(p^{2}+1\right)=(p-1)(p+1)\left(p^{2}+1\right)$.

Since $p$ is not divisible by three, either $p-1$ or $p+1$ is divisible by 3.

Consider the remainders of $p$ when divided by 5. If the remainder is 1, then $p-1$ is divisible by 5; if the remainder is 4, then $p+1$ is divisible by 5; and if the remainder is 2 or 3, then $p^{2}+1$ is divisible by 5.

Since $p$ is odd, each of the factors $p-1, p+1$, and $p^{2}+1$ is even. Moreover, one of the numbers $p-1$ or $p+1$ is divisible by 4, so the product of these three factors is divisible by 16.

It remains to show that if a number is greater than 240, it does not satisfy the condition. Note that the difference of the fourth powers of any prime numbers greater than 10 must be divisible by $n$. First, take the prime numbers 13 and 11 and compute the difference of their fourth powers: $13^{4}-11^{4}=(13-11)(13+11)\left(13^{2}+11^{2}\right)=2 \cdot 24 \cdot 290=240 \cdot 2 \cdot 29$. Now, take the numbers 17 and 11: $17^{4}-11^{4}=(17-11)(17+11)\left(17^{2}+11^{2}\right)=6 \cdot 28 \cdot 410=240 \cdot 7 \cdot 41$. Since the GCD $(240 \cdot 2 \cdot 29,240 \cdot 7 \cdot 41)=240$, $n$ cannot be greater than 240.

Grading criteria.

“+” A complete and well-reasoned solution is provided

“土” A generally correct reasoning is provided, containing minor gaps or inaccuracies

“-” The problem is not solved or solved incorrectly
9.6. Find the largest $n$ such that the sum of the fourth powers of any $n$ prime numbers greater than 10 is divisible by $n$.