For $k \le \frac{n}{2}$ the quotient is at least $2,$ so the remainder is at most $\min(k - 1,\ n - 2k);$ try $k$ near $\frac{n}{3}$

Show $F(3m) = m - 2,$ $F(3m+1) = m - 1,$ and $F(3m+2) = m,$ each achieved at $k = m + 1,$ then sum over the range

For $k \le \frac{n}{2}$ the quotient $\lfloor \frac{n}{k} \rfloor$ is at least $2,$ so the remainder satisfies $f(n, k) \le n - 2k$ as well as $f(n, k) \le k - 1.$ Write $n = 3m + r$ with $r \in \{0, 1, 2\}.$ Dividing by $k = m + 1$ gives quotient $2$ and remainder $m + r - 2,$ so $F(n) \ge m + r - 2.$ Conversely, for $k \ge m + 1,$ $f(n, k) \le n - 2k \le m + r - 2,$ and for smaller $k$ the bound $f(n, k) \le k - 1$ finishes the job: when $r = 2$ it gives at most $m$ for $k \le m + 1;$ when $r = 1$ it gives at most $m - 1$ for $k \le m;$ and when $r = 0$ it gives at most $m - 2$ for $k \le m - 1,$ while $k = m$ divides $3m$ exactly, leaving remainder $0.$ Hence $$\begin{aligned} F(3m) &= m - 2, \\ F(3m + 1) &= m - 1, \\ F(3m + 2) &= m. \end{aligned}$$

Grouping $n = 20, \ldots, 100$ as triples $3m - 1,$ $3m,$ $3m + 1$ for $m = 7, \ldots, 33$ (note $F(3m - 1) = F(3(m-1) + 2)$ $= m - 1$), each triple contributes $(m - 1) + (m - 2) + (m - 1)$ $= 3m - 4,$ so $$\begin{aligned} \tiny \sum_{n=20}^{100} F(n) &= \sum_{m=7}^{33} (3m - 4) \\ &= 3 \cdot \frac{(7 + 33) \cdot 27}{2} - 4 \cdot 27 \\ &= 1620 - 108 = 1512. \end{aligned}$$

The requested remainder is $512.$