For any set $S$, let $P(S)$ be its power set, the set of all of its subsets. Over all sets $A$ of $2015$ arbitrary finite sets, let $N$ be the maximum possible number of ordered pairs $(S,T)$ such that $S \in P(A), T \in P(P(A))$, $S \in T$, and $S \subseteq T$. (Note that by convention, a set may never contain itself.) Find the remainder when $N$ is divided by $1000.$

[i] Proposed by Ashwin Sah [/i]
1. **Understanding the Problem:**
   We need to find the maximum number of ordered pairs \((S, T)\) such that \(S \in P(A)\), \(T \in P(P(A))\), \(S \in T\), and \(S \subseteq T\). Here, \(A\) is a set of 2015 arbitrary finite sets.

2. **Analyzing the Conditions:**
   - \(S \in P(A)\) means \(S\) is a subset of \(A\).
   - \(T \in P(P(A))\) means \(T\) is a subset of the power set of \(A\).
   - \(S \in T\) means \(S\) is an element of \(T\).
   - \(S \subseteq T\) means \(S\) is a subset of \(T\).

3. **Maximizing the Number of Pairs:**
   To maximize the number of such pairs \((S, T)\), we need to maximize the number of elements of \(A\) that are also subsets of \(A\). Ideally, for all \(a \in A\), \(a \subseteq A\). This ensures that any \(S\) could be part of a pair \((S, T)\) as long as we select a suitable superset \(T\) that contains both \(S\) and the elements of \(S\).

4. **Constructing the Set \(A\):**
   There exists a set \(A\) such that \(a \in A\) implies \(a \subseteq A\). Using the set-theoretic definition of natural numbers, we can define \(A\) to be the set representing the natural number 2015. In this definition:
   - \(0 = \{\}\)
   - \(1 = \{0\}\)
   - \(2 = \{0, 1\}\)
   - \(\ldots\)
   - \(2015 = \{0, 1, 2, \ldots, 2014\}\)

   For all \(S \in A\), \(S \subseteq A\).

5. **Counting the Pairs \((S, T)\):**
   Let \(|S| = n\). For each \(S\), we construct \(T\) such that \(S \in T\) and \(S \subseteq T\). \(T\) must contain at least \(n+1\) subsets of \(A\) (including \(S\) and the elements of \(S\)). This leaves \(2^{2015} - n - 1\) subsets of \(A\) that \(T\) may or may not contain, giving \(2^{2^{2015} - n - 1}\) possible \(T\)'s.

6. **Summing Over All Possible \(S\):**
   There are \(\binom{2015}{n}\) subsets of size \(n\) of \(A\). We seek:
   \[
   \sum_{n=0}^{2015} \binom{2015}{n} 2^{2^{2015} - n - 1}
   \]

7. **Simplifying the Expression:**
   Using the binomial theorem:
   \[
   \sum_{n=0}^{2015} \binom{2015}{n} 2^{2^{2015} - n - 1} = 2^{2^{2015} - 1} \sum_{n=0}^{2015} \binom{2015}{n} 2^{-n}
   \]
   \[
   = 2^{2^{2015} - 1} \left(\frac{1}{2} + 1\right)^{2015}
   \]
   \[
   = 2^{2^{2015} - 1} \left(\frac{3}{2}\right)^{2015}
   \]
   \[
   = 2^{2^{2015} - 2016} \cdot 3^{2015}
   \]

8. **Finding the Remainder Modulo 1000:**
   We need to find \(2^{2^{2015} - 2016} \cdot 3^{2015} \mod 1000\). Since \(2^{2015}\) is extremely large, we focus on the powers of 3 modulo 1000:
   \[
   3^{2015} \mod 1000
   \]
   Using Euler's theorem, since \(\phi(1000) = 400\):
   \[
   3^{400} \equiv 1 \mod 1000
   \]
   \[
   2015 \equiv 15 \mod 400
   \]
   \[
   3^{2015} \equiv 3^{15} \mod 1000
   \]
   Calculating \(3^{15} \mod 1000\):
   \[
   3^5 = 243
   \]
   \[
   3^{10} = 243^2 = 59049 \equiv 49 \mod 1000
   \]
   \[
   3^{15} = 49 \cdot 243 = 11907 \equiv 907 \mod 1000
   \]

The final answer is \(\boxed{907}\)