Multiply by $6! = 720:$ the sum is an integer exactly when $720$ divides $n^6 + 6n^5 + 30n^4$ $+ 120n^3 + 360n^2$

Checking mod $2$ and mod $3$ forces $n$ to be a multiple of $6;$ write $n = 6k,$ and only the condition that $5$ divides $k(k + 1)$ survives

Multiplying by $6! = 720,$ the sum is an integer exactly when $$\begin{aligned} &n^6 + 6n^5 + 30n^4 \\ &{}+ 120n^3 + 360n^2 \\ &\text{is divisible by } 720. \end{aligned}$$ If $n$ were odd, every term except $n^6$ would be even, making the total odd. If $n$ is not divisible by $3,$ then modulo $3$ every term except $n^6$ vanishes while $n^6 \equiv 1 \pmod 3.$ So $n$ must be a multiple of $6.$

Write $n = 6k.$ Then $30n^4 = 720 \cdot 54k^4,$ $120n^3 = 720 \cdot 36k^3,$ and $360n^2 = 720 \cdot 18k^2$ are all divisible by $720,$ while $n^6 + 6n^5 = 6^6 k^5(k + 1).$ Since $6^6 = 2^6 3^6$ supplies the factors $2^4$ and $3^2$ of $720 = 2^4 \cdot 3^2 \cdot 5,$ the condition reduces to $k(k + 1)$ being divisible by $5,$ that is, $k \equiv 0$ or $4 \pmod 5.$

For $n = 6k \lt 2017$ we need $1 \le k \le 336.$ That range contains $67$ multiples of $5$ and $67$ values with $k \equiv 4 \pmod 5,$ so there are $67 + 67 = 134$ such $n.$