We wrote the numbers $1,2,3, \ldots, 20$ on the board. We erase two of them, $a$ and $b$, and write the number $a b+a+b$ in their place. After performing this procedure 19 times, what number can remain on the board?
Let $a \circ b = ab + a + b = (a+1)(b+1) - 1$; thus, $a \circ b$ represents the number we write on the board when $a$ and $b$ are erased. It is clear that this operation is commutative. We will show that $(a \circ b) \circ c = a \circ (b \circ c)$, so it is also associative:

$(a \circ b) \circ c = ((a+1)(b+1) - 1 + 1)(c+1) - 1 = (a+1)(b+1)(c+1) - 1, a \circ (b \circ c) = (a+1)((b+1)(c+1) - 1 + 1) - 1 = (a+$

This means that the final result does not depend on the order in which the numbers are erased. Next, we will show that for any $x_{1}, x_{2}, \ldots, x_{n}$ numbers,

$$
x_{1} \circ x_{2} \circ \cdots \circ x_{n} = (x_{1} + 1)(x_{2} + 1) \ldots (x_{n} + 1) - 1
$$

We will prove this by induction on $n$. We have just shown that the statement is true for $n \leq 3$. Assume that the statement is true for some $n$. Then, using this assumption,

$$
\begin{gathered}
x_{1} \circ x_{2} \circ \ldots x_{n} \circ x_{n+1} = \left((x_{1} + 1)(x_{2} + 1) \ldots (x_{n} + 1) - 1\right) \circ x_{n+1} = \\
= (x_{1} + 1)(x_{2} + 1) \ldots (x_{n} + 1)(x_{n+1} + 1) - 1
\end{gathered}
$$

so it is true for $(n+1)$ and therefore for all natural numbers. Therefore, the same number always remains on the board, which is

$$
(1+1) \cdot (2+1) \cdots \cdot (20+1) - 1 = 2 \cdot 3 \cdots \cdot 21 - 1 = 21! - 1
$$