An interesting problem with "decomposing" natural numbers.

Here’s a visual proof, to complement the algebra of other answers:

When you start the game (from 20), draw a “staircase” shape like in the figure, but with 19 squares in the base (so also 19 squares high). As you play, for each number on the board you’ll always have a corresponding staircase, with base and height 1 less than that number. Each turn, when you split up a number as $n = b+c$, split up its staircase as shown in the picture; that gives you a $b \times c$ rectangle, plus two smaller staircases for the resulting numbers $b$ and $c$. The area of the rectangles is your score so far. When all the numbers left are 1’s, then you’ve converted the whole original staircase into rectangles — so your final score is the total area of the original staircase.

This area, the number of squares in the staircase of base $n-1$, is given by the formula $\frac{n(n-1)}{2}$, as noted in other answers. This is a famous formula, and if you haven’t seen it before, it can be explained by the fact that two such staircases fit together into an $n \times (n-1)$ rectangle.

Suppose your hypothesis is that starting with $n$ you end up with a score of $\frac12 n(n-1)$. It is clearly true starting with $n=1$ as there are no moves and so a score of $0$.

Now suppose you know this is true for $1 \le n \le k$ for some $k$, then start with $k+1$ and split it into $a$ and $k+1-a$ where both are between $1$ and $k$. You get an immediate score of $a(k+1-a)$ plus (by the hypothesis) later scores of $\frac12 a(a-1)$ and $\frac12 (k+1-a)(k+1-a-1)$. Add these up and simplify to $\frac12 (k+1)k$. So it is true for $n=k+1$.

Using strong induction, you can conclude the hypothesis is true for all positive integers $n$.

Suppose that we represent a number $n$ on the whiteboard by $n$ distinct objects. When we split $a$ into $b+c$, we put $b$ of the objects in one group, and $c$ of the objects in the other group.

Then we can represent the $b\cdot c$ points we get for the split as follows: for every pair of objects that used to be in the same group, but are now in different groups, we get a point.

At the beginning, all $20$ objects are in the same group. At the end, all $20$ objects are in different groups, so we must have gotten $\binom{20}{2}$ points for separating them.