For each integer $1\le j\le 2017$, let $S_j$ denote the set of integers $0\le i\le 2^{2017} - 1$ such that $\left\lfloor \frac{i}{2^{j-1}} \right\rfloor$ is an odd integer. Let $P$ be a polynomial such that
\[P\left(x_0, x_1, \ldots, x_{2^{2017} - 1}\right) = \prod_{1\le j\le 2017} \left(1 - \prod_{i\in S_j} x_i\right).\]
Compute the remainder when
\[ \sum_{\left(x_0, \ldots, x_{2^{2017} - 1}\right)\in\{0, 1\}^{2^{2017}}} P\left(x_0, \ldots, x_{2^{2017} - 1}\right)\]
is divided by $2017$.

[i]Proposed by Ashwin Sah[/i]
1. **Define the sets \( S_j \):**
   For each integer \( 1 \le j \le 2017 \), the set \( S_j \) consists of integers \( 0 \le i \le 2^{2017} - 1 \) such that \( \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \) is an odd integer. This can be written as:
   \[
   S_j = \{ i \mid 0 \le i \le 2^{2017} - 1, \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \text{ is odd} \}
   \]

2. **Express the polynomial \( P \):**
   The polynomial \( P \) is given by:
   \[
   P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} \left(1 - \prod_{i \in S_j} x_i \right)
   \]
   Let \( T_j = \prod_{i \in S_j} x_i \). Then:
   \[
   P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} (1 - T_j)
   \]

3. **Evaluate the polynomial \( P \):**
   The polynomial \( P \) evaluates to 1 if all \( T_j \) are 0, and 0 if at least one \( T_j \) is 1. We need to count the number of tuples \( (x_0, x_1, \ldots, x_{2^{2017} - 1}) \in \{0, 1\}^{2^{2017}} \) such that all \( T_j \) are 0.

4. **Use complementary counting and the Principle of Inclusion-Exclusion (PIE):**
   Let \( A_j \) denote the set of tuples where \( T_j = 1 \). We want to compute:
   \[
   X = |P| - \sum |A_j| + \sum |A_i \cap A_j| - \sum |A_i \cap A_j \cap A_k| + \cdots
   \]
   where \( |P| = 2^{2^{2017}} \).

5. **Calculate the sizes of intersections:**
   - \( |A_j| = 2^{2^{2016}} \) because \( |S_j| = 2^{2016} \).
   - For \( |A_i \cap A_j| \) with \( i < j \), \( S_j \) fills half of the gaps in \( S_i \), so \( |A_i \cap A_j| = 2^{2^{2015}} \).
   - Generally, \( |A_{a_1} \cap A_{a_2} \cap \cdots \cap A_{a_k}| = 2^{2^{2017 - k}} \).

6. **Sum using PIE:**
   \[
   X = \sum_{k=0}^{2017} (-1)^k \binom{2017}{k} 2^{2^{2017 - k}}
   \]
   Simplifying modulo 2017:
   \[
   X \equiv 2^{2^{2017}} - 2 \pmod{2017}
   \]

7. **Compute \( 2^{2^{2017}} \mod 2017 \):**
   Using Fermat's Little Theorem, \( 2^{2016} \equiv 1 \pmod{2017} \). We need to find \( 2^{2017} \mod 2016 \):
   \[
   2^{2017} \equiv 2 \pmod{2016}
   \]
   Thus:
   \[
   2^{2^{2017}} \equiv 2^2 = 4 \pmod{2017}
   \]

8. **Final calculation:**
   \[
   X \equiv 4 - 2 = 2 \pmod{2017}
   \]

The final answer is \(\boxed{2}\)