Let $f, g$ be bijections on $\{1, 2, 3, \dots, 2016\}$. Determine the value of $$\sum_{i=1}^{2016}\sum_{j=1}^{2016}[f(i)-g(j)]^{2559}.$$
 
1. We start by considering the given double sum:
   \[
   S = \sum_{i=1}^{2016} \sum_{j=1}^{2016} [f(i) - g(j)]^{2559}
   \]
   Since \( f \) and \( g \) are bijections on the set \(\{1, 2, 3, \ldots, 2016\}\), we can reindex the sums using \( k = f(i) \) and \( l = g(j) \). This reindexing does not change the value of the sum because \( f \) and \( g \) are bijections, meaning they map each element of the set to a unique element of the set.

2. Rewriting the sum with the new indices:
   \[
   S = \sum_{k=1}^{2016} \sum_{l=1}^{2016} (k - l)^{2559}
   \]

3. Next, we observe the symmetry in the expression \((k - l)^{2559}\). Specifically, for any pair \((k, l)\), there is a corresponding pair \((l, k)\) such that:
   \[
   (k - l)^{2559} = -(l - k)^{2559}
   \]
   Since \(2559\) is an odd number, \((k - l)^{2559} = -(l - k)^{2559}\).

4. We can pair each term \((k - l)^{2559}\) with its corresponding term \((l - k)^{2559}\) in the sum:
   \[
   S = \sum_{k=1}^{2016} \sum_{l=1}^{2016} (k - l)^{2559}
   \]
   For each pair \((k, l)\) and \((l, k)\), the terms \((k - l)^{2559}\) and \((l - k)^{2559}\) will cancel each other out because:
   \[
   (k - l)^{2559} + (l - k)^{2559} = (k - l)^{2559} - (k - l)^{2559} = 0
   \]

5. Therefore, the entire sum \(S\) is zero because every term cancels out with its corresponding term:
   \[
   S = 0
   \]

The final answer is \(\boxed{0}\).