KVPY #1

$\begin{aligned} N & = \frac {1^3+2^3+3^3+\cdots + (2n)^3}{1^2+2^2+3^2+\cdots + n^2} \\ & = \frac {\left(\frac {2n(2n+1)}2 \right)^2}{\frac {n(n+1)(2n+1)}6} \\ & = \frac {6n^2(2n+1)^2}{n(n+1)(2n+1)} \\ & = \frac {6n(2n+1)}{n+1} \\ & = \frac {6n(2n+2)-6n}{n+1}\\ & = \frac {12n(n+1)-6(n+1)+6}{n+1} \\ & = 12n - 6 + \frac 6{n+1} \end{aligned}$

For $N$ to be an integer, $\dfrac 6{n+1}$ has to be an integer and there are only three cases of positive integer $n=1, 2, 5$ . And, their sum is $1+2+5= \boxed{8}$ .

After solving the complex above we get

12n^2+6n/n+1

6n(n+1)+6n^2/n+1

(6n)+(6n^2)/n+1

(6n(n+1)-6n)/n+1

6n/n+1

(6(n+1)-6)/6

6/n+1

n=1,2,5

N=1+2+5=8

$\dfrac{1^3 + 2^3 + \cdots + (2n)^3 }{1^2 + 2^2 + \cdots + n^2} = \dfrac{\sum_{k=1}^{2n} k^3}{\sum_{k=1}^{n} k^2} = \dfrac{\frac{{(2n)}^2 {(2n+1)}^2}{4} }{\frac{n (n+1) (2n+1)}{6}} = \dfrac{6n(2n+1)}{n+1}$

Now for $\dfrac{6n(2n+1)}{n+1} \in \mathbb{Z}^{+}$ , three situations are possible

$\mathbf{(I)} \ n+1 \ | \ 6 \implies n \in \{1,2,5\}$

$\mathbf{(II)} \ n+1 \ | \ n \implies n \in \emptyset$

$\mathbf{(III)} \ n+1 \ | \ 2n+1 \implies n+1 \ | \ n + (n+1) \implies n+1 \ | \ n \implies n \in \emptyset$

Thus the sum of all values of $n$ is $1+2+5 = \boxed{8}$ .