Given that $\{a_n\}$ is a sequence of integers satisfying the following condition for all positive integral values of $n$: $a_n+a_{n+1}=2a_{n+2}a_{n+3}+2016$. Find all possible values of $a_1$ and $a_2$
1. **Claim:** \(a_1 = a_3\) and \(a_2 = a_4\).

   **Proof:**
   Let \(s_n = a_{2n+2} - a_{2n}\). Notice that:
   \[
   a_{2n} + a_{2n+1} = 2a_{2n+2}a_{2n+3} + 2016
   \]
   \[
   a_{2n+1} + a_{2n+2} = 2a_{2n+3}a_{2n+4} + 2016
   \]
   Taking the difference of these two equations yields:
   \[
   (a_{2n+1} + a_{2n+2}) - (a_{2n} + a_{2n+1}) = 2a_{2n+3}a_{2n+4} - 2a_{2n+2}a_{2n+3}
   \]
   Simplifying, we get:
   \[
   a_{2n+2} - a_{2n} = 2a_{2n+3}(a_{2n+4} - a_{2n+2})
   \]
   Therefore:
   \[
   s_n = 2a_{2n+3}s_{n+1}
   \]
   Now suppose on the contrary that \(s_n > 0\) for all \(n\). Then \(|a_{2n+3}| \geq 1\) for all \(n\). Hence:
   \[
   |s_n| = 2|a_{2n+3}||s_{n+1}| > |s_{n+1}|
   \]
   This implies that \(\{s_n\}\) is a strictly decreasing sequence of positive integers, which is a contradiction.

   Therefore, \(s_n = 0\) for some \(n \in \mathbb{N}\). By the above inequality, we have \(s_{n+1} = 0\) implies \(s_n = 0\). So by backward induction, we have \(s_1 = 0\). Therefore, \(a_2 = a_4\) and by symmetry \(a_1 = a_3\). \(\blacksquare\)

2. Therefore:
   \[
   a_1 + a_2 = 2a_3a_4 + 2016 = 2a_1a_2 + 2016
   \]
   Simplifying, we get:
   \[
   a_1 + a_2 = 2a_1a_2 + 2016
   \]
   \[
   a_1 + a_2 - 2a_1a_2 = 2016
   \]
   \[
   2a_1a_2 - a_1 - a_2 = -2016
   \]
   Adding 1 to both sides:
   \[
   2a_1a_2 - a_1 - a_2 + 1 = -2015
   \]
   Factoring, we get:
   \[
   (2a_1 - 1)(2a_2 - 1) = 4031
   \]
   We have \(4031 = 29 \times 139\). Therefore, the possible pairs \((2a_1 - 1, 2a_2 - 1)\) are:
   \[
   (-4031, -1), (-139, -29), (-29, -139), (-1, -4031), (1, 4031), (29, 139), (139, 29), (4031, 1)
   \]
   Solving for \(a_1\) and \(a_2\), we get:
   \[
   (a_1, a_2) = (0, 2016), (-14, 70), (-69, 15), (-2015, 1), (2016, 0), (70, -14), (15, -69), (1, -2015)
   \]

3. Conversely, suppose \((a_1, a_2)\) are the values above. Then define \(a_{2n} = a_2\) and \(a_{2n+1} = a_1\). We easily see that it satisfies the requirements.

The final answer is \( \boxed{ (a_1, a_2) = (0, 2016), (-14, 70), (-69, 15), (-2015, 1), (2016, 0), (70, -14), (15, -69), (1, -2015) } \).