What we want are $c = 1, 1008, 1009, 2016$, and the number of $f$ is $2017$.
Let $a\%b$ denote the remainder of $a$ divided by $b$, where $0 \le a\%b < b$.
Note that $(cx + k)\%2017$ satisfies the condition (where $k = 0, 1, \dots, 2016$).

In the following, we'll prove there is another solution when $c \ne 1, 1008, 1009, 2016$:
When $c \ne 0$, let
$$f_0(2017n + 1) = f_0(2017n + c + 1) = f_0(2017n + 2c + 1) = c,$$
let $f_0(x) = 0$ for the others. $(f_0(x) + f_0(y))\%2017$ can only be $0, c, 2c\%2017$,
but since $c \ne 1, 1008, 1009, 2016$, we have
$0, c, 2c\%2017 \notin \{1, (c+1)\%2017, (2c+1)\%2017\}$,
so
$f_0(f_0(x) + f_0(y)) = 0$, and $f_0(f_0(x) + f_0(y) + 1) = c$.
So $f_0$ satisfies the condition, and $f_0$ is different from $(cx + k)\%2017$.

Let $f_0(x) = 1$ for the others. ($f_0(x) + f_0(y)\%2017$ can only be $2, 4, 6$, and
$$f_0(2) = f_0(3) = f_0(4) = f_0(5) = f_0(6) = f_0(7),$$
so $f_0$ satisfies the condition, and $f_0$ is different from $(cx + k)\%2017$.

Next, we'll prove when $c = 1, 1008, 1009, 2016$, all solutions are $(cx + k)\%2017$:
Assume $f$ is a solution satisfying the condition, we'll prove $f$ is onto first. If $f$ is not onto, we can find a longest "good sequence" $\{a_i\}_{i=1}^r$ in the range of $f$, where "good sequence" means a sequence in the form $a_i = (ci + d)\%2017$. Since $2017$ is a prime, and $f$ is not onto, the longest good sequence we found has finite length.

Substitute $(x, y)$ into (2) such that $(f(x), f(y)) = (a_i, a_j), i, j = 1, 2, \dots, r$:
When $c = 1$, let $b_i = f(a_1 + a_i), b_{r+1} = f(a_1 + a_r + 1)$.
When $c = 1008$, let $b_i = f(a_{r+1-i} + a_{r+1-i}), b_{r+1} = f(2a_1 + 1)$.
When $c = 1009$, let $b_i = f(a_i + a_i), b_{r+1} = f(2a_r + 1)$.
When $c = 2016$, let $b_i = f(a_1 + a_{r+1-i}), b_{r+1} = f(2a_1 + 1)$.
Then we have $\{b_i\}_{i=1}^{r+1}$ as a longer good sequence, contradiction! So $f$ is onto, and we get $f(x+1) - f(x) \equiv c \pmod{2017}$, $\forall x$. So there exists $k$ such that $f(x) = (cx + k)\%2017$.