jmc

Notice that ${\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$, so$$\begin{array}{rl}\sum _{i=n+1}^{2n}{i}^2& =\sum _{i=1}^{2n}{i}^2-\sum _{i=1}^n{i}^2\\ & =\frac{2n(2n+1)(4n+1)}{6}-\frac{n(n+1)(2n+1)}{6}\\ & =\frac{16n^3+12n^2+2n}{6}-\frac{2n^3+3n^2+n}{6}\\ & =\frac{14n^3+9n^2+n}{6}\\ & =\frac{n(2n+1)(7n+1)}{6}\end{array}$$Thus, $\left({\sum }_{i=1}^n{i}^2\right)\left({\sum }_{i=n+1}^{2n}{i}^2\right)=\frac{n^2(2n+1{)}^2(n+1)(7n+1)}{36}$. In order for the expression to be a perfect square, $(n+1)(7n+1)$ must be a perfect square. By using the Euclidean Algorithm, $\gcd (n+1,7n+1)=\gcd (n+1,6)$. Thus, the GCD of $n+1$ and $7n+1$ must be factors of 6. Now, split the factors as different casework. Note that the quadratic residues of 7 are 0, 1, 2, and 4. If $\gcd (n+1,7n+1)=6$, then $n\equiv 5(\mathrm{mod}6)$. Let $n=6a+5$, so $(n+1)(7n+1)=(6a+6)(42a+36)=36(a+1)(7a+6)$. Since 6 is divided out of $n+1$ and $7n+1$, $a+1$ and $7a+6$ are relatively prime, so $a+1$ and $7a+6$ must be perfect squares. However, since 6 is not a quadratic residue of 7, the GCD of $n+1$ and $7n+1$ can not be 6. If $\gcd (n+1,7n+1)=3$, then $n\equiv 2(\mathrm{mod}3)$. Let $n=3a+2$, so $(n+1)(7n+1)=(3a+3)(21a+15)=9(a+1)(7a+5)$. Since 3 is divided out of $n+1$ and $7n+1$, $a+1$ and $7a+5$ are relatively prime, so $a+1$ and $7a+5$ must be perfect squares. However, since 5 is not a quadratic residue of 7, the GCD of $n+1$ and $7n+1$ can not be 3. If $\gcd (n+1,7n+1)=2$, then $n\equiv 1(\mathrm{mod}2)$. Let $n=2a+1$, so $(n+1)(7n+1)=(2a+2)(14a+8)=4(a+1)(7a+4)$. Since 2 is divided out of $n+1$ and $7n+1$, $a+1$ and $7a+4$ are relatively prime, so $a+1$ and $7a+4$ must be perfect squares. We also know that $n+1$ and $7n+1$ do not share a factor of 3, so $n\equiv 1,3(\mathrm{mod}6)$. That means $n\le 2007$, so $a\le 1003$. After trying values of $a$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $a=960$. That means $n=1921$. If $\gcd (n+1,7n+1)=1$, then $n+1\equiv 1,5(\mathrm{mod}6)$ (to avoid common factors that are factors of 6), so $n\equiv 0,4(\mathrm{mod}6)$. After trying values of $n$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $n=120$ (we could also stop searching once $n$ gets below 1921). From the casework, the largest natural number $n$ that makes $(1^2+2^2+3^2+\cdots +n^2)\left[(n+1{)}^2+(n+2{)}^2+(n+3{)}^2+\cdots +(2n{)}^2\right]$ is a perfect square is $\boxed{1921}$.