This is really a simple problem, isn't it!

Let the two perfect squares be $a-b$ and $a+b$ . Then

$(a+b)^2 - (a-b)^2 = 4ab = 4004 \Rightarrow ab = 1001.$

$n = \frac{(a+b)^2+(a-b)^2}{2} = a^2 + b^2.$

We must therefore consider all possible factorizations $a\cdot b = 1001$ ; each of them contributes $a^2 + b^2$ to the final sum. In other words, the answer is simply the sum of squares of all divisors of 1001:

$N = \sum_{d\|1001} d^2,$

Since $1001 = 7\times 11\times13$ , there are eight divisors, which can be factored as:

$d = (1 \text{ or } 7)\cdot(1 \text{ or } 11)\cdot(1 \text{ or } 13).$

Thus the sum of the squares of these divisors is

$\sum (1+7^2)\cdot(1+11^2)\cdot(1+13^2) = 50\times 122\times 170 = 1037000.$

$\begin{cases} a^2=n+2002 \\ b^2=n-2002 \end{cases} \Leftrightarrow \begin{cases} a^2-b^2=4004 \\ n=\frac{a^2+b^2}{2} \end{cases} \quad \Leftrightarrow (a+b)\cdot(a-b)=4004$

Now, intending a clearer approach, let´s make

$\begin{cases} x=a+b \\ y=a-b \end{cases} \quad \Leftrightarrow \begin{cases} a=\frac{x+y}{2} \\ b=\frac{x-y}{2} \end{cases}$

Then,

$x\cdot y=13\times11\times7\times2\times2\times1=4004$

We can just think in the values for x and y that give us positives integers values for a and b, because for n, a and b must be integers and they are squared.

By the factorization we just have integers values for a and b if x and y are both even, which give us,

1) $x=2002; y=2 \rightarrow a=1002; b=1000 \quad \rightarrow \boxed{n=1002002}$

$\rightarrow \sqrt{n+2002}=\sqrt{1004004}=1002 \rightarrow \sqrt{n-2002}=\sqrt{1000000}=1000$

2) $x=286; y=14 \rightarrow a=150; b=36 \quad \rightarrow \boxed{n=20498}$

$\rightarrow \sqrt{n+2002}=\sqrt{22500}=150 \rightarrow \sqrt{n-2002}=\sqrt{18496}=136$

3) $x=182; y=22 \rightarrow a=102; b=80 \quad \rightarrow \boxed{n=8402}$

$\rightarrow \sqrt{n+2002}=\sqrt{10404}=102 \rightarrow \sqrt{n-2002}=\sqrt{6400}=80$

4) $x=154; y=26 \rightarrow a=90; b=64 \quad \rightarrow \boxed{n=6098}$

$\rightarrow \sqrt{n+2002}=\sqrt{8100}=90 \rightarrow \sqrt{n-2002}=\sqrt{4096}=64$

So,

$\sum{n}=1002002+20498+8402+6098=1037000$

$\square$ Reference

It is an utter pleasure add a solution, even that just to a Level 4 problem. From a Totally Level 5 Member. Satyajit Mohanty. The Mathematical Nerd.

Let n - 2002 = k^2, n + 2002 = l^2; subtracting, 4004 = l^2 - k^2 = (l - k)(l + k). 4004 = 2^2 7 11 13. We seek factors l - k and l + k which are both even. Clearly, both cannot be odd, and they cannot be of opposite parity, for their sum = 2l would be odd, a contradiction. Alittle reflection shows that the only products available are (1): 2 2002, (2) 14 286, (3) 22 182, and (4) 26*154.In each case, setting l - k = smaller factor, l + k = larger factor, and subtracting, dividing by 2, we have l = 1002, 150, 102, and 90. For each value of l, we compute l^2 - 2002 = n, and sum the four answers getting 1037000. Ed Gray