p is odd prime. Let’s note that 1^7+...+p^7 is divisible by p (because a^7 is equivalent to -(p-a)^7, so their sum is divisible by p). So, 1^7+...+(pk)^7 is divisible by p. Then let’a note that \frac{n(n-1)}{2}-1= \frac{n(n-1)-2}{2} Numerator of this fraction is divisible by n-2. If n-2 has odd prime divisor p, then 1^7+...+n^7 is divisible by p and as I noted above 1^7+...+(n-2)^7 is divisible by p. So, (n-1)^7+n^7 is divisible by p, that is 1+2^7 is divisible by p. So, p is either 3 or 43. It means that n-2 can has only 2,3 and 43 as prime divisors. Moreover, n-2 cannot be divisible by 9 or 43^2 (because sum of a^7 and (p^2-a)^7 gives 0 mod p^2, so 1^7+...+(p^2k)^7 is divisible by p^2, so 129 is divisible by either 43^2 or 9, we got contradiction). If n-2 is divisible by 4, then \frac{n(n-1)-2}{2} is even, but 1^7+...+n^7 is odd. Hence, all prime divisors of n-2 may be 2, 3,43. And we need to consider 7 cases: n-2=2 \frac{n(n-1)-2}{2}=5 1^7+...+4^7 is divisible by 5 n-2=3 \frac{n(n-1)-2}{2}=9 1^7+...+5^7 isn’t divisible by 9 n-2=6 \frac{n(n-1)-2}{2}=27 1^7+...+5^7+6^7+7^7 isn’t divisible by 9 n-2=43 \frac{n(n-1)-2}{2}=989 1^7+...+45^7 is divisible by 23 and 43 n-2=86 \frac{n(n-1)-2}{2}=3827 1^7+...+88^7 is divisible by 43 and 89 n-2=129 \frac{n(n-1)-2}{2}=8514 1^7+...+131^7 isn’t divisible by 9 n-2=258 \frac{n(n-1)-2}{2}=33669 1^7+...+260^7 isn’t divisible by 27 So, answer is n=4, 45, 88.

pixi wrote: p is odd prime. Let’s note that $1^7+...+p^7 $ is divisible by $p$ (because $a^7$ is equivalent to $-(p-a)^7$, so their sum is divisible by $p$). So, $1^7+...+(pk)^7$ is divisible by $p$. Then let’a note that $\frac{n(n-1)}{2}-1= \frac{n(n-1)-2}{2}$ Numerator of this fraction is divisible by $n-2$. If $n-2$ has odd prime divisor $p$, then $1^7+...+n^7$ is divisible by $p $ and as I noted above $1^7+...+(n-2)^7$ is divisible by $p$. So, $(n-1)^7+n^7$ is divisible by $p$, that is $1+2^7$ is divisible by $p$. So, $p$ is either $3$ or $43$. It means that $n-2$ can has only $2,3$ and $43$ as prime divisors. Moreover, $n-2$ cannot be divisible by $9$ or $43^2$ (because sum of $a^7$ and $(p^2-a)^7$ gives $0$ mod $p^2$, so $1^7+...+(p^2k)^7$ is divisible by $p^2$, so $129$ is divisible by either $43^2$ or $9$, we got contradiction). If $n-2$ is divisible by $4$, then $\frac{n(n-1)-2}{2}$ is even, but $1^7+...+n^7$ is odd. Hence, all prime divisors of $n-2$ may be $2, 3,43$. And we need to consider $7$ cases: $n-2=2$ $\frac{n(n-1)-2}{2}=5$ $1^7+...+4^7$ is divisible by $5$ $n-2=3$ $\frac{n(n-1)-2}{2}=9$ $1^7+...+5^7$ isn’t divisible by $9$ $n-2=6$ $\frac{n(n-1)-2}{2}=27$ $1^7+...+5^7+6^7+7^7$ isn’t divisible by $9$ $n-2=43$ $\frac{n(n-1)-2}{2}=989$ $1^7+...+45^7$ is divisible by $23$ and $43$ $n-2=86$ $\frac{n(n-1)-2}{2}=3827$ $1^7+...+88^7$ is divisible by $43$ and $89$ $n-2=129$ $\frac{n(n-1)-2}{2}=8514$ $1^7+...+131^7$ isn’t divisible by $9$ $n-2=258$ $\frac{n(n-1)-2}{2}=33669$ $1^7+...+260^7$ isn’t divisible by $27$ So, answer is $n=4, 45, 88$. Here is the awesome solution of pixi just $\LaTeX$-ed

Anyway here is my bashy solution. We use Faulhaber's theorem to get that: $$\sum_{t=1}^{n} t^7 = \frac{3(n(n+1))^4-4(n(n+1))^3+2(n(n+1))^2}{24}$$ Thus the equivalent problem is the following: $$12(n+1)(n-2) \mid ((n+1)n)^2(3(n(n+1))^2-4n(n+1)+2)$$$$12(n-2) \mid (n+1)n^2(3(n(n+1))^2-4n(n+1)+2)$$ Now to solve this we waste an hour on a bash through all of the remainders for $12$. That means that we iterate through every $r$ in $n=12k+r$. (I know pretty ugly), and we get the solutions $n\in \{4,45,88\}$

Why not n=1 and 3 are not solutions.

The answer is $n\in\{1, 3, 4, 8, 45, 88, 260\}$. Let $S(n) = \sum_{i=1}^n {i^7}$. $\binom{1}{2} - 1 = -1 \mid 1^7$, so $1$ is a solution. If $n>2$ is odd, $\binom{n}{2} - 1 \equiv \frac{2\cdot 1}{2} - 1 \equiv 0 \pmod {n-2}$. Hence, we must have (using the well-known fact that $k\mid S(k)\forall k>2$) $$S(n) = S(n-2) + (n-1)^7 + n^7 \equiv 1^7 + 2^7 \equiv 129 \equiv 0 \pmod {n-2}$$which implies $n\in \{3, 5, 45, 131\}$. Checking, $n=3$ and $n=45$ are solutions. If $n = 4k+2$, $\binom{n}{2}-1\equiv 0 \pmod 2$, but $2 \nmid S(n)$, so we get no solutions. If $n = 4k$, $\binom{n}{2} - 1\equiv 0 \pmod {2k-1}$. Thus $$S(4k) = S(4k-2) + (4k-1)^7 + (4k)^7 \equiv 1^7 + 2^7 \equiv 129 \equiv 0 \pmod {2k-1}$$So $n\in \{4, 8, 88, 260\}$. Checking, all of them are solutions.