Problem 1203

1. **Determine the constraints on $n$ being even:**
- Since $n$ is even, we consider the highest power of 2 dividing $n$. Let $v_2(n)$ denote the highest power of 2 dividing $n$.
- Suppose $v_2(n) \geq 4$. Then $n$ is divisible by 16. We need $m^6 \equiv 1 \pmod{16}$ for all $m$ coprime to 16.
- Consider $m = 5$, which is coprime to 16. We have:
$$5^6 = 15625 \equiv 9 \pmod{16}$$
- This is a contradiction since $5^6 \not\equiv 1 \pmod{16}$. Therefore, $v_2(n) \leq 3$.

2. **Check $n$ being divisible by 8:**
- If $v_2(n) = 3$, then $n$ is divisible by 8. We need $m^6 \equiv 1 \pmod{8}$ for all $m$ coprime to 8.
- For any $m$ coprime to 8, $m$ is odd. We have:
$$m^2 \equiv 1 \pmod{8} \implies m^6 = (m^2)^3 \equiv 1^3 = 1 \pmod{8}$$
- Thus, $m^6 \equiv 1 \pmod{8}$ holds for all $m$ coprime to 8.

3. Check divisibility by other primes:
- Suppose $5 \mid n$. We need $m^6 \equiv 1 \pmod{5}$ for all $m$ coprime to 5.
- Consider $m = 3$, which is coprime to 5. We have:
$$3^6 = 729 \equiv 4 \pmod{5}$$
- This is a contradiction since $3^6 \not\equiv 1 \pmod{5}$. Therefore, $n$ cannot be divisible by 5.

4. **Determine the maximum $n$ with the given constraints:**
- Since $n$ must be coprime to 5 and divisible by 8, we consider the prime factors of $n$ other than 2 and 5.
- Let $n = 2^3 \cdot 3^2 \cdot 7$. We need to check if $m^6 \equiv 1 \pmod{n}$ for all $m$ coprime to $n$.
- For $n = 504$:
$$504 = 2^3 \cdot 3^2 \cdot 7$$
- We need $m^6 \equiv 1 \pmod{504}$ for all $m$ coprime to 504.
- By the Chinese Remainder Theorem, it suffices to check $m^6 \equiv 1 \pmod{8}$, $m^6 \equiv 1 \pmod{9}$, and $m^6 \equiv 1 \pmod{7}$.

5. Verify the conditions:
- For $m$ coprime to 8, $m^6 \equiv 1 \pmod{8}$ holds as shown earlier.
- For $m$ coprime to 9, $m^6 \equiv 1 \pmod{9}$ holds because the order of any element modulo 9 divides $\phi(9) = 6$.
- For $m$ coprime to 7, $m^6 \equiv 1 \pmod{7}$ holds because the order of any element modulo 7 divides $\phi(7) = 6$.

Thus, $n = 504$ satisfies all the conditions.

The final answer is $\boxed{504}$