Let $m$ be a positive integer less than $2015$. Suppose that the remainder when $2015$ is divided by $m$ is $n$. Compute the largest possible value of $n$.

[i]  Proposed by Michael Ren [/i]
1. We start with the given equation for division with remainder:
   \[
   2015 = qm + n
   \]
   where \( q \) is the quotient, \( m \) is the divisor, and \( n \) is the remainder. By definition, \( 0 \leq n < m \).

2. To maximize \( n \), we need to consider the largest possible value of \( n \) under the constraint \( n < m \). The largest possible value for \( n \) is \( n = m - 1 \).

3. Substituting \( n = m - 1 \) into the equation, we get:
   \[
   2015 = qm + (m - 1)
   \]
   Simplifying, we have:
   \[
   2015 = qm + m - 1 \implies 2016 = m(q + 1)
   \]

4. To find the maximum possible value of \( m \), we need to find the largest divisor of 2016 that is less than 2015. We start by factoring 2016:
   \[
   2016 = 2^5 \times 3^2 \times 7
   \]

5. The largest divisor of 2016 that is less than 2015 is 1008. We can verify this by checking the division:
   \[
   2016 = 1008 \times 2
   \]
   Thus, \( m = 1008 \).

6. Substituting \( m = 1008 \) back into the equation for \( n \):
   \[
   n = m - 1 = 1008 - 1 = 1007
   \]

Therefore, the largest possible value of \( n \) is \( \boxed{1007} \).