1. Divide the numbers $1,2, \cdots, 20$ into two groups, Jia and Yi, such that the average of the numbers in group Jia is 1 more than the average of the numbers in group Yi. Then group Jia has ( ) numbers.
(A) 8
(B) 9
(C) 10
(D) 11
$-1 . C$.
Let group A have $a$ numbers. Then group B has $20-a$ numbers. Let the average of the numbers in group B be $b$. Then the average of the numbers in group A is $b+1$. According to the problem,
$$
a(b+1)+b(20-a)=1+2+\cdots+20 \text {. }
$$

Simplifying the above equation gives $a=210-20 b$.
Since $0<a<20$, we have
$$
0<210-20 b<20 \text {. }
$$

Solving this gives $9 \frac{1}{2}<b<10 \frac{1}{2}$.
Therefore, $b=10, a=10$.
We can take the numbers in group A to be: $1,7,8,9,10,11,12,13$,
$$
19,20 \text {; }
$$

The numbers in group B are: $2,3,4,5,6,14,15,16,17,18$.