I was considering the sequence of pairs
$$(1,100), (2,99), (3,98), (4,97) ... $$
To try to find a pair of positive integers $(K, 100-k)$ such that $\text{GCD}(K, 101-K)$ is not $1$.
And after running a short python script I came to the conclusion that NO such pair exists.
I wondered if this had something to do with $101$ being a prime number so I explored the sequence of pairs $(K,M-K)$ where $M$ was prime.
$$(1,2) \ M=3$$ $$(1,4), (2,3) \ M=5$$ $$(1,6), (2,5), (3,4) \ M= 7$$ $$(1,10), (2,9), (3,8), (4,7), (5,6) \ M=11 $$ $$ (1,12), (2,11), (3,10), (4,9), (5,8), (6,7) \ M = 13 $$
To confirm this isn't just "odd numbers" $M=9$ fails, so it feels like i'm onto something.
Some more experimental computation with a python script, confirmed it for primes up to $47$.
So this should have some elementary explanation i'm thinking, but i'm not sure what it could be.
If $M$ is not prime, then let $1 \neq d \mid M$. Since $1 < d < M$, $(d, M - d)$ is one of such pairs which have $d$ as a common divisor.
If $M$ is prime, then suppose there exists $k \in \{1,\dots,M - 1\}$ in which $\gcd(k, M - k) = d > 1$. Then $d \mid (M - k) + k \Rightarrow d \mid M$, so $d = M$ as $d > 1$. This is a contradiction, as $k < M$ yet $M \mid k$.