Let $p=2017$ be a prime.  Find the remainder when \[\left\lfloor\dfrac{1^p}p\right\rfloor + \left\lfloor\dfrac{2^p}p\right\rfloor+\left\lfloor\dfrac{3^p}p\right\rfloor+\cdots+\left\lfloor\dfrac{2015^p}p\right\rfloor \] is divided by $p$.  Here $\lfloor\cdot\rfloor$ denotes the greatest integer function.

[i]Proposed by David Altizio[/i]
To solve the problem, we need to find the remainder when the sum 

\[
\left\lfloor \frac{1^p}{p} \right\rfloor + \left\lfloor \frac{2^p}{p} \right\rfloor + \left\lfloor \frac{3^p}{p} \right\rfloor + \cdots + \left\lfloor \frac{2015^p}{p} \right\rfloor 
\]

is divided by \( p = 2017 \). Here, \( \lfloor \cdot \rfloor \) denotes the greatest integer function.

1. **Using Fermat's Little Theorem:**
   Fermat's Little Theorem states that for any integer \( a \) and a prime \( p \), we have:
   \[
   a^p \equiv a \pmod{p}
   \]
   This implies:
   \[
   \frac{a^p}{p} = \frac{a}{p} + \frac{kp}{p} = \frac{a}{p} + k
   \]
   where \( k \) is an integer.

2. **Applying the Floor Function:**
   For \( a \) in the range \( 1 \leq a \leq 2016 \), we have:
   \[
   \left\lfloor \frac{a^p}{p} \right\rfloor = \left\lfloor \frac{a}{p} + k \right\rfloor = k
   \]
   Since \( \frac{a}{p} \) is less than 1 for \( a < p \), \( k \) is the integer part of \( \frac{a^p}{p} \).

3. **Pairing Terms:**
   Consider the pairs \( (x, p-x) \) for \( x = 1, 2, \ldots, 1008 \). For each pair:
   \[
   \left\lfloor \frac{x^p}{p} \right\rfloor + \left\lfloor \frac{(p-x)^p}{p} \right\rfloor
   \]
   Using Fermat's Little Theorem:
   \[
   x^p \equiv x \pmod{p} \quad \text{and} \quad (p-x)^p \equiv p-x \pmod{p}
   \]
   Thus:
   \[
   \frac{x^p}{p} + \frac{(p-x)^p}{p} = \frac{x}{p} + \frac{p-x}{p} = 1
   \]
   Since \( \frac{x}{p} \) and \( \frac{p-x}{p} \) are not integers, we use the lemma:
   \[
   \left\lfloor \frac{x^p}{p} \right\rfloor + \left\lfloor \frac{(p-x)^p}{p} \right\rfloor = 1 - 1 = 0
   \]

4. **Summing the Pairs:**
   There are 1007 pairs, each contributing 0 to the sum. Additionally, we have:
   \[
   \left\lfloor \frac{1^p}{p} \right\rfloor = \left\lfloor \frac{1}{2017} \right\rfloor = 0
   \]

5. **Final Calculation:**
   The sum of all terms is:
   \[
   0 + 0 + \cdots + 0 = 0
   \]
   The remainder when 0 is divided by 2017 is:
   \[
   0
   \]

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