Let $$S_K = \dfrac{1+2+\ldots+K}{K}$$ and $$\sum_{j=1}^n S_j^2 = \dfrac{n}{A}(Bn^2 + Cn + D)$$ where $$A, B, C, D \in N$$ and $$A$$ has least value, then

We start by computing $$S_K$$ as the average of the first $$K$$ integers: $$S_K = \frac{1+2+\ldots+K}{K} = \frac{K(K+1)/2}{K} = \frac{K+1}{2}$$.

Next we form the sum of squares $$\sum_{j=1}^n S_j^2$$. Substituting the formula for $$S_j$$ gives $$\sum_{j=1}^n S_j^2 = \sum_{j=1}^n \frac{(j+1)^2}{4} = \frac{1}{4}\sum_{j=1}^n (j+1)^2 = \frac{1}{4}\sum_{m=2}^{n+1} m^2$$ by letting $$m=j+1$$.

Using the identity $$\sum_{m=1}^M m^2 = \frac{M(M+1)(2M+1)}{6}$$ with $$M = n+1$$ and subtracting the first term gives $$\frac{1}{4}\Bigl(\frac{(n+1)(n+2)(2n+3)}{6}-1\Bigr) = \frac{1}{4}\cdot\frac{(n+1)(n+2)(2n+3)-6}{6}$$.

Expanding the cubic in the numerator, $$(n+1)(n+2)(2n+3) = 2n^3 + 9n^2 + 13n + 6$$, leads to $$\frac{1}{4}\cdot\frac{2n^3 + 9n^2 + 13n + 6 - 6}{6} = \frac{2n^3 + 9n^2 + 13n}{24} = \frac{n(2n^2 + 9n + 13)}{24}$$.

From this final form we set $$A = 24$$, $$B = 2$$, $$C = 9$$ and $$D = 13$$.

Then $$A + B = 24 + 2 = 26$$, and since $$26/13 = 2$$, it follows that $$A + B$$ is divisible by $$D$$.