Suppose $P(x)$ is a quadratic polynomial with integer coefficients satisfying the identity \[P(P(x)) - P(x)^2 = x^2+x+2016\] for all real $x$.  What is $P(1)$?
1. Let \( P(x) = ax^2 + bx + c \). Given the identity:
   \[
   P(P(x)) - P(x)^2 = x^2 + x + 2016
   \]
   we need to substitute \( P(x) \) into the equation and simplify.

2. First, compute \( P(P(x)) \):
   \[
   P(P(x)) = P(ax^2 + bx + c) = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c
   \]

3. Next, compute \( P(x)^2 \):
   \[
   P(x)^2 = (ax^2 + bx + c)^2
   \]

4. Substitute these into the given identity:
   \[
   a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c - (ax^2 + bx + c)^2 = x^2 + x + 2016
   \]

5. Simplify the left-hand side:
   \[
   (a-1)(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c = x^2 + x + 2016
   \]

6. Since the identity must hold for all \( x \), compare the coefficients of the powers of \( x \) on both sides. Start with the highest power, \( x^4 \):
   \[
   (a-1)a^2 = 0 \implies a = 1 \quad \text{(since } a \neq 0 \text{ for a quadratic polynomial)}
   \]

7. Substitute \( a = 1 \) into the equation:
   \[
   b(x^2 + bx + c) + c = x^2 + x + 2016
   \]

8. Expand and compare coefficients:
   \[
   b(x^2 + bx + c) + c = bx^2 + b^2x + bc + c = x^2 + x + 2016
   \]

9. Compare the coefficients of \( x^2 \):
   \[
   b = 1
   \]

10. Compare the coefficients of \( x \):
    \[
    b^2 = 1 \implies b = 1 \quad \text{(since } b \text{ is already determined as 1)}
    \]

11. Compare the constant terms:
    \[
    bc + c = 2016 \implies 2c = 2016 \implies c = 1008
    \]

12. Thus, the polynomial \( P(x) \) is:
    \[
    P(x) = x^2 + x + 1008
    \]

13. Finally, compute \( P(1) \):
    \[
    P(1) = 1^2 + 1 + 1008 = 1 + 1 + 1008 = 1010
    \]

The final answer is \( \boxed{1010} \).