Plusses and minuses are inserted in the expression \[\pm 1 \pm 2 \pm 3 \dots \pm 2016\] such that when evaluated the result is divisible by 2017. Let there be $N$ ways for this to occur. Compute the remainder when $N$ is divided by 503.

[i]2016 CCA Math Bonanza Team #10[/i]
1. **Define the problem in terms of subsets:**
   We need to find the number of ways to insert pluses and minuses in the expression \(\pm 1 \pm 2 \pm 3 \dots \pm 2016\) such that the result is divisible by 2017. This is equivalent to finding subsets of \(\{1, 2, 3, \ldots, 2016\}\) such that the sum of the elements in the subset is \(0 \mod 2017\).

2. **Define the function \(f\):**
   Let \(f(x)\) be the function that returns the number of subsets of \(\{1, 2, 3, \ldots, 2016\}\) such that the sum of the elements in the subset is \(x \mod 2017\).

3. **Sum of all subsets:**
   Clearly, the total number of subsets of \(\{1, 2, 3, \ldots, 2016\}\) is \(2^{2016}\). Therefore, we have:
   \[
   f(0) + f(1) + f(2) + \cdots + f(2016) = 2^{2016}
   \]

4. **Establishing the bijection:**
   For any non-zero \(i\), \(f(i) = f(i+1)\). This is because given a subset \(S\) that sums to \(i\), we can always find a subset that sums to \(i+1\) by either adding or removing an element. This establishes a bijection between subsets summing to \(i\) and those summing to \(i+1\).

5. **Special case for \(f(0)\):**
   The only exception is the subset \(\{1, 2, 3, \ldots, 2016\}\) itself, which sums to \(\frac{2016 \cdot 2017}{2} = 2016 \cdot 1008 = 2032128 \equiv 0 \mod 2017\). Therefore, \(f(0) = f(1) + 1\).

6. **Solving for \(f(i)\):**
   Let \(f(i) = x\) for \(i \neq 0\). Then:
   \[
   f(0) = x + 1
   \]
   \[
   f(0) + 2016x = 2^{2016}
   \]
   Substituting \(f(0) = x + 1\):
   \[
   (x + 1) + 2016x = 2^{2016}
   \]
   \[
   2017x + 1 = 2^{2016}
   \]

7. **Taking modulo 503:**
   Since \(2017 \equiv 1 \mod 503\), we have:
   \[
   2017x + 1 \equiv 2^{2016} \mod 503
   \]
   We need to compute \(2^{2016} \mod 503\). Using Fermat's Little Theorem, since 503 is prime:
   \[
   2^{502} \equiv 1 \mod 503
   \]
   \[
   2^{2016} = (2^{502})^4 \cdot 2^8 \equiv 1^4 \cdot 2^8 \equiv 2^8 \mod 503
   \]
   \[
   2^8 = 256
   \]
   Therefore:
   \[
   2017x + 1 \equiv 256 \mod 503
   \]
   \[
   2017x \equiv 255 \mod 503
   \]
   \[
   x \equiv 255 \mod 503
   \]

8. **Finding \(f(0)\):**
   \[
   f(0) = x + 1 \equiv 255 + 1 \equiv 256 \mod 503
   \]

The final answer is \(\boxed{256}\).