1. Find $S_K$

Given

$$S_K=\frac{1+2+\cdots+K}{K}$$

Using

$$1+2+\cdots+K=\frac{K(K+1)}{2}$$

we get

$$S_K=\frac{\frac{K(K+1)}{2}}{K}=\frac{K+1}{2}$$

So,

$$S_j^2=\left(\frac{j+1}{2}\right)^2=\frac{(j+1)^2}{4}$$

2. Compute the sum

We need

$$\sum_{j=1}^{n} S_j^2=\sum_{j=1}^{n}\frac{(j+1)^2}{4}=\frac14\sum_{j=1}^{n}(j+1)^2$$

Now,

$$\sum_{j=1}^{n}(j+1)^2=\sum_{j=1}^{n}(j^2+2j+1)$$

Hence,

$$\sum_{j=1}^{n}(j+1)^2=\sum_{j=1}^{n}j^2+2\sum_{j=1}^{n}j+\sum_{j=1}^{n}1$$

Using standard formulas,

$$\sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}{6}, \qquad \sum_{j=1}^{n}j=\frac{n(n+1)}{2}, \qquad \sum_{j=1}^{n}1=n$$

So,

$$\sum_{j=1}^{n}(j+1)^2=\frac{n(n+1)(2n+1)}{6}+n(n+1)+n$$

Therefore,

$$\sum_{j=1}^{n} S_j^2 =\frac14\left[\frac{n(n+1)(2n+1)}{6}+n(n+1)+n\right]$$

Take $n$ common:

$$=\frac{n}{4}\left[\frac{(n+1)(2n+1)}{6}+(n+1)+1\right]$$

Simplify inside the bracket:

$$\frac{(n+1)(2n+1)}{6}+(n+1)+1 =\frac{(n+1)(2n+1)+6(n+1)+6}{6}$$ $$=(2n^2+3n+1+6n+6+6)/6 =\frac{2n^2+9n+13}{6}$$

Hence,

$$\sum_{j=1}^{n} S_j^2=\frac{n}{4}\cdot \frac{2n^2+9n+13}{6} =\frac{n(2n^2+9n+13)}{24}$$

So it is of the form

$$\frac{n}{A}(Bn^2+Cn+D)$$

with

$$A=24,\quad B=2,\quad C=9,\quad D=13$$

Since $\gcd(24,2,9,13)=1$, $A=24$ is already least.

3. Check each option

Option A

$$A+B+C+D=24+2+9+13=48$$

$48$ is not divisible by $5$.

So, A is false.

Option B

$$A+C+D=24+9+13=46$$

Check divisibility by $B=2$:

$$46 \text{ is divisible by } 2$$

But the statement says "not divisible".

So, B is false.

Option C

Left side:

$$A+B=24+2=26$$

Right side:

$$5(D-C)=5(13-9)=5\cdot 4=20$$

Since $26\neq 20$,

So, C is false.

Option D

$$A+B=24+2=26$$

Check divisibility by $D=13$:

$$26 \text{ is divisible by } 13$$

So, D is true.

4. Conclusion

The correct option is

$$\boxed{\text{D}}$$

5. Comparison with stored answer

Stored correct answer: D

Our derived answer matches the stored answer.