3. Let the set $X=\{1,2, \cdots, 20\}, A$ be a subset of $X$, the number of elements in $A$ is at least 2, and all elements of $A$ can be arranged as consecutive positive integers. Then the number of such sets $A$ is $\qquad$.
If $|A|=2$, then there are 19 sets $A$; if $|A|=3$, then there are 18 sets $A$; $\cdots$; 
if $|A|=20$, then there is 1 set $A$. Therefore, the number of sets $A$ that satisfy the condition is 
$$
1+2+\cdots+19=\frac{19 \times(1+19)}{2}=190 \text {. }
$$