$A=A( m,n) =\sum ^{m+n-1}_{k=m} k^{2} /n=m^{2} +( n-1) m+\left( 2n^{2} -3n+1\right) /6$. First, we will show that if $ n$ satisfies the condition, then $ n\equiv \pm 1,\pm 7\bmod 24$. Suppose that $ n$ satisfies the condition, i.e. $ A$ is a perfect square for some $ m\in \mathbb{Z}$. Since $ 3\mid 2n^{2} -3n+1$, we must have $ 3\nmid n$. Assume that $ n\equiv 3\bmod 8$. Since $ 2n^{2} -3n+1\equiv 2\bmod 8$, we have $ \left( 2n^{2} -3n+1\right) /6\equiv 3\bmod 4$. So we have $ A\equiv m^{2} +2m+3\equiv ( m+1)^{2} +2\equiv 2,3\bmod 4$, which is impossible. Assume that $ n\equiv 5\bmod 8$. Since $ 2n^{2} -3n+1\equiv 4\bmod 8$, we have $ \left( 2n^{2} -3n+1\right) /6\equiv 2\bmod 4$. So we have $ A\equiv m^{2} +4m+2\equiv ( m+2)^{2} -2\equiv 2,3\bmod 4$, which is impossible. So we must have $ n\equiv \pm 1\bmod 8$. Therefore, we have $ n\equiv \pm 1,\pm 7\bmod 24.$ Second, we will show that if $ n\equiv \pm 1,\pm 7\bmod 24$, then $ n$ satisfies the condition. We consider the following four cases $ ( 1) ,( 2) ,( 3) ,( 4)$. $ ( 1)$ $ n=24t+1$ for $ t\in \mathbb{N}_{\geq 0}$. Then $ A\left( 12t^{2} -11t-1,24t+1\right) =\left( 12t^{2} +t+1\right)^{2}$. $ ( 2)$ $ n=24t-1$ for $ t\in \mathbb{N}$. Then $ A\left( 12t^{2} -13t,24t-1\right) =\left( 12t^{2} -t+1\right)^{2}$. $ ( 3)$ $ n=24t+7$ for $ t\in \mathbb{N}_{\geq 0}$. Then $ A\left( 12t^{2} -5t-3,24t+7\right) =\left( 12t^{2} +7t+2\right)^{2}$. $ ( 4)$ $ n=24t-7$ for $ t\in \mathbb{N}$. Then $ A\left( 12t^{2} -19t+4,24t-7\right) =\left( 12t^{2} -7t+2\right)^{2}$. Hence the answer is $ n\equiv \pm 1,\pm 7\bmod 24$.

. . so lets just write every thing for $m>0$ first . we want $[(m+1)^2+...+(n+m)^2]/n$ be a square which is equal to $[(m+n)^3+3(m+n)^2+(m+n)-m^3-3m^2-m]/6n$ . which is $ m^2+(n+1)m+ [(2n^2+3n+1)/6]$ so we must have $n=6t+1$ or $n=6t-1$ lets check the first one , we have $n=6t+1$ so we must have $(m+3t+1)^2 + (12t^2+7t+1-9t^2-6t-1)$ is a perfect square lets have $m+3t+1=s$ and then there should exist $r$ such that $3t^2+t=r^2+2rs$ by looking mode$2$ we have that $2|r$ so let $r=2$ so $3t^2+t=4(s+1)$ and we should just have $4|t^2-t$ so $t=4k+1,4k$ or $n=24k+7,24k+1$but remember that we $SHOULD$ have $12t+8<3t^2+t$ which is true for $t>4$ these cases have been dealt with now , for finishing let $n=6t-1$ we have that $m^2+6tm+12t^2-t$ is a perfect square letting $m+3t=r$ we should have $r^2+3t^2-t=v^2$ so there exists $s$ such that $3t^2-t=s^2+2sr$ again by checking mode $2$ we get $2|s$ and again we have a solution for $24t-7,24t-1$ but we also should have $3t^2+t>12t$ which is true for $t>3$ . $m\le0$ any way here is the rest . lets have $m=0$ then we should have $(n-1)(2n-1)=6s^2$ since $(n-1,2n-1)=1$ we have $n-1=2s^2,2n-1=3r^2$ $n-1=6s^2,2n-1=r^2$ in the forst case we have $4s^2+1=3r^2$ which is impossible so $12s^2+1=r^2$ which is $r^2-12s^2=1$ a pell equation with the base answer of $r=7,s=2$ and all of the solutions form these forms are attainable now for $m<0$ let $|m|=d$ we want $1+4+...+d^2+1+...+(n-d-1)^2=nk^2$ so lets write everything again , we must have $d(d+1)(2d+1)+(n-d-1)(n-d)(2n-2d-1)=6nk^2$ so $2d^3+3d^2+d+2(n-d)^3-3(n-d)^2+(n-d)=6nk^2$ which is aswell $2n^3-3n^2+n+dn(-6n+6d+6)=6nk^2$ devide by $6m$ to get $(2n^2-3n+1)/6 + d^2-dn+d=k^2$ so again we must have $ n=6t+1,6t-1$ lets go to check the first one , we must have $3t^2+t+(d-3t)^2 =k^2$ and again there are $u,v$ such that $3t^2+t=u^2+2uv$ cheching mode $2$ we have $2|u$ let $u=2$ and again the same happens which im too lazy to write anymore . and the solutions are $n=24t+1,24t+7,24t-1,24t-7$ as kaede mentioned ... infact we really didnt need to seperate the problem in cases but anywas , that was my solution ...