Let $\lfloor x \rfloor$ denote the largest integer less than or equal to $x$, and let $\{x\}$ denote the fractional part of $x$. For example, $\lfloor \pi \rfloor=3$, and $\{\pi\}=0.14159\dots$, while $\lfloor 100 \rfloor=100$ and $\{100\}=0$. If $n$ is the largest solution to the equation $\frac{\lfloor n \rfloor}{n}=\frac{2015}{2016}$, compute $\{n\}$. 
1. We start with the given equation:
   \[
   \frac{\lfloor n \rfloor}{n} = \frac{2015}{2016}
   \]
   Multiplying both sides by \( n \) gives:
   \[
   \lfloor n \rfloor = \frac{2015}{2016} n
   \]

2. Let \( n = \lfloor n \rfloor + \{ n \} \), where \( \lfloor n \rfloor \) is the integer part of \( n \) and \( \{ n \} \) is the fractional part of \( n \). Substituting this into the equation, we get:
   \[
   \lfloor n \rfloor = \frac{2015}{2016} (\lfloor n \rfloor + \{ n \})
   \]

3. Distribute \(\frac{2015}{2016}\):
   \[
   \lfloor n \rfloor = \frac{2015}{2016} \lfloor n \rfloor + \frac{2015}{2016} \{ n \}
   \]

4. Rearrange to isolate \(\lfloor n \rfloor\):
   \[
   \lfloor n \rfloor - \frac{2015}{2016} \lfloor n \rfloor = \frac{2015}{2016} \{ n \}
   \]
   \[
   \lfloor n \rfloor \left(1 - \frac{2015}{2016}\right) = \frac{2015}{2016} \{ n \}
   \]
   \[
   \lfloor n \rfloor \left(\frac{1}{2016}\right) = \frac{2015}{2016} \{ n \}
   \]

5. Multiply both sides by \( 2016 \):
   \[
   \lfloor n \rfloor = 2015 \{ n \}
   \]

6. Since \(\lfloor n \rfloor\) is an integer and \(0 \leq \{ n \} < 1\), \(\{ n \}\) can be expressed as \(\frac{k}{2015}\) for some integer \(k\) where \(0 \leq k \leq 2014\). This implies:
   \[
   n = \lfloor n \rfloor + \frac{k}{2015}
   \]

7. Substituting \(\lfloor n \rfloor = k\) into the equation, we get:
   \[
   n = k + \frac{k}{2015} = \frac{2016k}{2015}
   \]

8. To find the largest solution, we set \(k\) to its maximum value, which is 2014:
   \[
   n = \frac{2016 \times 2014}{2015}
   \]

9. Simplify the expression:
   \[
   n = \frac{2016 \times 2014}{2015} = 2014 + \frac{2014}{2015}
   \]

10. Therefore, the fractional part of \( n \) is:
    \[
    \{ n \} = \frac{2014}{2015}
    \]

The final answer is \(\boxed{\frac{2014}{2015}}\).