On $[k, k+1)$ with $k \ge 0$ an integer, write $x = k + t;$ then $f(x) = (k + t)t$ is strictly increasing from $0$ toward $k + 1,$ so $f$ maps $[k, k+1)$ bijectively onto $[0, k+1).$ Hence for any $c$ with $n \lt c \lt n + 1,$ the equation $f(w) = c$ has exactly one solution in $[k, k+1)$ for each integer $k \ge n,$ and no others.

The equation $f(z) = 17$ has one solution $z_n \in (n, n+1)$ for each $n \ge 17.$ In turn, $f(w) = z_n$ has one solution in $(k, k+1)$ for each $k \ge n.$ Finally, for $x \in [j, j+1)$ with $0 \le j \le 2019,$ $f$ maps $[j, j+1)$ bijectively onto $[0, j+1),$ so the number of solutions of $f(f(f(x))) = 17$ there equals the number of such $w$ with $w \lt j + 1,$ namely the number of pairs $(n, k)$ with $17 \le n \le k \le j,$ which is $\binom{j - 15}{2}.$ (The endpoint $x = 2020$ gives $f(x) = 0$ and is not a solution.)

By the hockey stick identity, $$\begin{aligned} N &= \sum_{j=17}^{2019} \binom{j - 15}{2} \\ &= \sum_{a=2}^{2004} \binom{a}{2} \\ &= \binom{2005}{3} \\ &= \frac{2005 \cdot 2004 \cdot 2003}{6} \\ &= 1{,}341{,}349{,}010, \end{aligned}$$ so the remainder when $N$ is divided by $1000$ is $10.$