You've got a good intuition, but how to prove it ? Here's what I'd go with, it's voluntarily very detailed :
For all $A \in \mathbb{N}$, we'll be looking for $x, y \in \left[1; A-1\right]$ so that
$$
\cases{0<x<y \\ x+y = A}
$$
And we'll try to determine how many values can $x$ take. If we know this, then we know how many combinations there are.
Let's start with finding the maximum value for $x$.
If $x=\frac{A+1}2$, then :
$$
x+y>x+x=A+1>A
$$
So this $x$ does not fit our case. Therefore we must have $x<\frac{A+1}2$. Since $x$ is a natural integer, this is equivalent to $x\le\frac{A+1}2-1=\frac{A-1}2$.
We have a maximum bound $M=\frac{A-1}2$.
Now, does it work for every integer $x$ between $1$ and $\left\lfloor M\right\rfloor$ ?
It's pretty obvious that for any integer $x \in \left[1; \left\lfloor M\right\rfloor\right]$, there exists another integer $y \in \left[\left\lfloor M\right\rfloor+1; A-1\right]$ so that $x+y=A$, and $x \ne y$
Let's see for the limit cases :
- $x=1$ : choose $y=A-1$, and $x+y=1+(A-1)=A$
- $x=2$ : choose $y=A-2$, and $x+y=2+(A-2)=A$
- ...
- $x=\left\lfloor M\right\rfloor-1$ : choose $y=\left\lceil M\right\rceil+2$, and $x+y=2M+1=(A-1)+1=A$
- $x=\left\lfloor M\right\rfloor$ : choose $y=\left\lceil M\right\rceil+1$, and $x+y=2M+1=(A-1)+1=A$
We have found that all $x$ between $1$ and $M$ works, and that $x$ cannot be lower than $1$ or greater than $M$. Thus we have found that there are exactly $\left\lfloor M\right\rfloor=\left\lfloor\frac{A-1}2\right\rfloor$ valid combinations.
For $A=100$, this gives :
$$
\left\lfloor M\right\rfloor=\left\lfloor\frac{100-1}2\right\rfloor=\left\lfloor49.5\right\rfloor=49
$$