Rewrite the sum as ${\sum }_{r=1}^n(x+2r-2)(x+2r)=\frac{8n}{3}$.

Expanding each term and summing the standard $\sum r$ and $\sum r^2$ pieces:

$n{x}^2+2x{\sum }_{r=1}^n(2r-1)+4{\sum }_{r=1}^{n}r(r-1)=\frac{8n}{3}$

$n{x}^2+2x\cdot n^2+\frac{4n(n^2-1)}{3}-\frac{8n}{3}=0$

Dividing by $n$ gives a quadratic in $x$: $x^2+2nx+\frac{4(n^2-1)}{3}-\frac{8}{3}=0$.

For two consecutive even-integer roots $\alpha ,\beta$ we need $|\alpha -\beta |=2$, i.e. $\frac{\sqrt{D}}{|a|}=2\Rightarrow D=4$.

Here $D=4n^2-4\left(\frac{4(n^2-1)}{3}-\frac{8}{3}\right)=4$, which simplifies to $n^2-\frac{4n^2}{3}=-3$, giving $n^2=9$, so $n=3$.