For positive integers $n$ and $k$, let $f(n, k)$ be the remainder when $n$ is divided by $k$, and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$. Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$.
The Pattern
We can find that 
$20\equiv 6 \pmod{7}$
$21\equiv 5 \pmod{8}$
$22\equiv 6 \pmod{8}$
$23\equiv 7 \pmod{8}$
$24\equiv 6 \pmod{9}$
$25\equiv 7 \pmod{9}$
$26\equiv 8 \pmod{9}$
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
$99\equiv 31 \pmod{34}$
$100\equiv 32 \pmod{34}$
So the sum is $6+3\times(6+...+31)+31+32=1512$,it is also 17+20+23+...+95, so the answer is $\boxed{512}$.
By: Kris17

The Intuition
First, let's see what happens if we remove a restriction. Let's define $G(x)$ as
$G(x):=\max_{\substack{1\le k}} f(n, k)$
Now, if you set $k$ as any number greater than $n$, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence $H(x)$ is defined as
$H(x):=\max_{\substack{1\le k\le n}} f(n, k)$
Now, after some thought, we find that if we set $k=\lfloor \frac{n}{2} \rfloor+1$ we get a remainder of $\lfloor \frac{n-1}{2} \rfloor$, the max possible. Once we have gotten this far, it is easy to see that the original equation, 
$F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$
has a solution with $k=\lfloor \frac{n}{3} \rfloor+1$.
$W^5$~Rowechen

The Proof
The solution presented above does not prove why $F(x)$ is found by dividing $x$ by $3$. Indeed, that is the case, as rigorously shown below.
Consider the case where $x = 3k$. We shall prove that $F(x) = f(x, k+1)$.
For all $x/2\ge n > k+1, x = 2n + q$, where $0\le q k+1$.
Consider all $n  5, f(x, k+1) > f(x, n)$ for $n < k+1$.
Similar proofs apply for $x = 3k + 1$ and $x = 3k + 2$. The reader should feel free to derive these proofs themself.

Generalized Solution
$Lemma:$ Highest remainder when $n$ is divided by $1\leq k\leq n/2$ is obtained for $k_0 = (n + (3 - n \pmod{3}))/3$ and the remainder thus obtained is $(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]$. 
$Note:$ This is the second highest remainder when $n$ is divided by $1\leq k\leq n$ and the highest remainder occurs when $n$ is divided by $k_M$ = $(n+1)/2$ for odd $n$ and  $k_M$ = $(n+2)/2$ for even $n$.
Using the lemma above:
$\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}]$
$= [(120*81/2)/3 - 2*81 + (2/3)*81] = 1512$
So the answer is $\boxed{512}$
Proof of Lemma: It is similar to $The Proof$ stated above. 
Kris17