The squares of two positive integers differ by 2016. Find the maximum possible sum of the two integers.

[i]Proposed by Clive Chan
1. We start with the given equation:
   \[
   x^2 - y^2 = 2016
   \]
   This can be factored using the difference of squares:
   \[
   (x-y)(x+y) = 2016
   \]

2. To find integer solutions for \(x\) and \(y\), both \(x-y\) and \(x+y\) must be integers. Additionally, since \(x\) and \(y\) are both integers, \(x-y\) and \(x+y\) must have the same parity (both even or both odd).

3. We need to find pairs \((a, b)\) such that \(a = x-y\) and \(b = x+y\) and \(ab = 2016\). We also need \(a\) and \(b\) to be of the same parity.

4. First, we factorize 2016:
   \[
   2016 = 2^5 \times 3^2 \times 7
   \]
   We list all factor pairs of 2016:
   \[
   (1, 2016), (2, 1008), (3, 672), (4, 504), (6, 336), (7, 288), (8, 252), (9, 224), (12, 168), (14, 144), (16, 126), (18, 112), (21, 96), (24, 84), (28, 72), (32, 63), (36, 56), (42, 48)
   \]

5. We need to find pairs where both numbers are of the same parity:
   - \( (2, 1008) \)
   - \( (6, 336) \)
   - \( (8, 252) \)
   - \( (12, 168) \)
   - \( (14, 144) \)
   - \( (18, 112) \)
   - \( (24, 84) \)
   - \( (28, 72) \)
   - \( (36, 56) \)
   - \( (42, 48) \)

6. For each pair \((a, b)\), we solve for \(x\) and \(y\):
   \[
   x = \frac{a + b}{2}, \quad y = \frac{b - a}{2}
   \]
   We need to ensure that both \(x\) and \(y\) are integers.

7. We check the pairs:
   - For \((2, 1008)\):
     \[
     x = \frac{2 + 1008}{2} = 505, \quad y = \frac{1008 - 2}{2} = 503
     \]
     Sum: \(505 + 503 = 1008\)

   - For \((6, 336)\):
     \[
     x = \frac{6 + 336}{2} = 171, \quad y = \frac{336 - 6}{2} = 165
     \]
     Sum: \(171 + 165 = 336\)

   - For \((8, 252)\):
     \[
     x = \frac{8 + 252}{2} = 130, \quad y = \frac{252 - 8}{2} = 122
     \]
     Sum: \(130 + 122 = 252\)

   - For \((12, 168)\):
     \[
     x = \frac{12 + 168}{2} = 90, \quad y = \frac{168 - 12}{2} = 78
     \]
     Sum: \(90 + 78 = 168\)

   - For \((14, 144)\):
     \[
     x = \frac{14 + 144}{2} = 79, \quad y = \frac{144 - 14}{2} = 65
     \]
     Sum: \(79 + 65 = 144\)

   - For \((18, 112)\):
     \[
     x = \frac{18 + 112}{2} = 65, \quad y = \frac{112 - 18}{2} = 47
     \]
     Sum: \(65 + 47 = 112\)

   - For \((24, 84)\):
     \[
     x = \frac{24 + 84}{2} = 54, \quad y = \frac{84 - 24}{2} = 30
     \]
     Sum: \(54 + 30 = 84\)

   - For \((28, 72)\):
     \[
     x = \frac{28 + 72}{2} = 50, \quad y = \frac{72 - 28}{2} = 22
     \]
     Sum: \(50 + 22 = 72\)

   - For \((36, 56)\):
     \[
     x = \frac{36 + 56}{2} = 46, \quad y = \frac{56 - 36}{2} = 10
     \]
     Sum: \(46 + 10 = 56\)

   - For \((42, 48)\):
     \[
     x = \frac{42 + 48}{2} = 45, \quad y = \frac{48 - 42}{2} = 3
     \]
     Sum: \(45 + 3 = 48\)

8. The maximum possible sum of the two integers is \(1008\).

The final answer is \(\boxed{1008}\).