OMO 2012

Let $x_{1} + x_{2} + ... + x_{2011} + x_{2012}$ be some permutation of the list

I deduced a greddy aproach to perform the operations and get that the maximum computation of N (I'll put a proof of this later).

$N = (\sum_{i=1}^{2011} 2^{i} \times x_{i}) + 2^{2011} * x_{2012}$

The permutation of the list that maximizes $N$ is one such that for every element $x_{i}$ , $x_{i} = i$

So let's rewritte N as $N = (\sum_{i=1}^{2011} 2^{i} \times {i}) + 2^{2011} * 2012$

Now here is a little bit of Javascript to make the computation.

(function(n, mod) {
  var pw = 1, ans = 0;

  for (var i=1; i<n; ++i) {
    pw = (pw * 2) % mod;
    ans = (ans + pw * i) % mod;
  }

  ans = (ans + pw * n) % mod;

  console.log(ans);
}).call(this, 2012, 1000);

The code bellow prints out the requested answer wich is $\boxed{538}$