Solution:

Let $N$ be the number whose base-$2016$ representation is $A B C$. That is,
$$N = A \times 2016^2 + B \times 2016 + C$$
where $A, B, C$ are digits in base $2016$, i.e., $0 \leq A, B, C \leq 2015$.

We are to compute the remainder when $N - (A+B+C+k)$ is divided by $2015$, for $k \in \{1,2, \ldots, 2015\}$.

First, note that $2016 = 2015 + 1$.

Let us compute $N$ modulo $2015$:

$2016 \equiv 1 \pmod{2015}$, so $2016^2 \equiv 1^2 = 1 \pmod{2015}$.

Therefore,
$$N \equiv A \times 1 + B \times 1 + C \pmod{2015}$$
So $N \equiv A + B + C \pmod{2015}$.

Therefore,
$$N - (A+B+C+k) \equiv (A+B+C) - (A+B+C) - k \equiv -k \pmod{2015}$$
So the remainder when $N - (A+B+C+k)$ is divided by $2015$ is $2015 - k$, unless $k = 2015$, in which case the remainder is $0$.

Final answer: $2015 - k$.