Perfect Square.

Suppose that $\begin{aligned} & n-50 = p^2 \\& n+50 = q^2 \end{aligned}$ Subtracting 2nd equation from first we get as $\begin{aligned} & q^2 -p^2 = 100 \\& (q+p)(q-p) = 100\end{aligned}$ The left sides of above equation is the product of all possible divisor of 100. So 100 has the following total productial divisors. $(q+p)(q-p) = \begin{cases} 1\times 100 \\ 2\times 50 \\ 4\times 25 \\ 5\times 20 \\10\times 10 \end{cases}$ Equating the LHS divisors to RHS one $p$ and $q$ will be only integer iff $(q-p)(p+q) = \begin{cases} 2\times 50 \\ 10\times 10\end{cases}$ Solving above 4 equations we get $q = 26, 24$ and $p= 10, 0$ and replugging the values in the above equations we get either $n =626$ or $n=50$

So the final answer is $\boxed{676}$ .