Super Quadratic!

$\begin{aligned} x(x+2)+(x+1)(x+2)+\cdots +(x+(n-1))(x+n) & = 10n \\ \implies \sum_{k=1}^n (x+(k-1))(x+k) & = 10n \\ \sum_{k=1}^n \left(x^2+(2k-1)x+k^2 -k \right) & = 10n \\ nx^2+(n(n+1)-n)x+\frac {n(n+1)(2n+1)}6 - \frac {n(n+1)}2 & = 10n & \small \color{#3D99F6} \text{Dividing both sides by }n \\ x^2+nx+\frac {(n+1)(2n+1)}6 - \frac {n+1}2 & = 10 \\ x^2+nx+\frac {(n+1)(2n+1-3)}6 & = 10 \\ x^2+nx+\frac {(n+1)(n-1)}3 & = 10 \\ x^2+nx+\frac {n^2-1}3 - 10 & = 0 \end{aligned}$

$\begin{aligned} \implies x & = \frac {-n \pm \sqrt{n^2 -\frac {4(n^2-1)}3+40}}2 = \frac {-n \pm \sqrt{\frac {4 - n^2}3+40}}2 \end{aligned}$

We note that $\sqrt{\frac {4 - n^2}3+40}$ is real for $1 \le n \le 11$ and for $x$ to be an integer, $\dfrac {4-n^2}3+40$ must be a perfect square. Let $m^2 = \dfrac {4-n^2}3+40$ , where $m$ is a positive integer. We note that $m^2<40$ and they are 36, 25, 16...; $n= \sqrt{3(40-m^2)+4}$ ; the roots are $\dfrac {-n+m}2, \dfrac {-n-m}2$ and integral roots occur when:

\(\begin{array} {} m^2=36 & \implies n = 4 & \implies x = -5, 1 \\ m^2=25 & \implies n = 7 & \implies x = -6, -1 \\ m^2=1 & \implies n = 11 & \implies x = -6, -5 \end{array} \)

Therefore, the answer is $\boxed{11}$ .