1. Find all positive integers $k$ such that the set $X=\{1994,1997$, $2000, \cdots, 1994+3k\}$ can be partitioned into two disjoint subsets $A$ and $B$ whose union is $X$, and the sum of the elements in $A$ is 9 times the sum of the elements in $B$. (1st Test, 2nd Question, provided by Huang Yiguo)
Let $X$ contain $k+1$ positive integers, the sum of these $k+1$ positive integers is denoted as $S$. According to the problem,
$$
\begin{aligned}
S & =1994+1997+2000+\cdots+(1994+3 k) \\
& =1990(k+1)+\frac{1}{2}(k-1)(3 k+13) .
\end{aligned}
$$

Let the new positive integer in $B$ be $x$, then the sum of all positive integers in $A$ is $9x$, thus:
$$
10 x=1990(k+1)+\frac{1}{2}(k+1)(3 k+8),
$$

Therefore,
$$
x=199(k+1)+\frac{1}{20}(k+1)(3 k+8) .
$$

From (2), we know that $(k+1)(3 k+8)$ must be a multiple of 20. Thus, there is only one possibility: $k=5 m-1(m \in \mathbb{N})$, and $m(15 m+5)$ is a multiple of 4. Therefore, $m=4 t$ or $m=4 t+1(t \in \mathbb{N})$.

Case 1: $m=4 t, k=20 t-1(t \in \mathbb{N})$. $X$ contains $20 t$ positive integers. Arrange the numbers in $X$ in ascending order, starting from 1994, take the maximum and minimum of every twenty consecutive positive integers and assign them to $B$, and the middle eighteen numbers to $A$. This way, the resulting $A$ and $B$ clearly meet the problem's requirements.

Case 2: $m=4 t+1, k=20 t+4(t \in \mathbb{N})$. Using (2) again, we have
$$
x=5(4 t+1)(3 t+200) .
$$

Since each positive integer in $X$ leaves a remainder of 2 when divided by 3, let $|B|$ denote the number of positive integers in set $B$. Then, combining (3), we have
$$
2|B|=x(\bmod 3) \equiv t+1(\bmod 3) .
$$

Thus,
$$
|B| \equiv 2 t+2(\bmod 3) .
$$

When $t=7$, from (3), we have
$$
x=1990 \times 13+6175=1990 \times 16+205 \text{. }
$$

Here, the sum of the largest 13 positive integers in $X$ is
$$
(1990+436)+(1990+433)+\cdots+(1990+400)
$$
$$
=1990 \times 13+418 \times 13=1990 \times 16+205 .
$$

From (5), (6), (7), and (8), it is clear that when $t=7$, there does not exist a decomposition of $X$ into subsets $A$ and $B$ that meets the problem's requirements.
When $t=8$, $x=1990 \times 18+1140$.

Let
$$
B=\{1994,1997,2000, \cdots, 2036,2039,2210,
$$
$2486\}$ (a total of 18 positive integers).
The sum of all positive integers in $B$ is exactly $1990 \times 18+1140$. Let $A=X-B$, then when $t=8$, the set $X$ has a decomposition that meets the problem's requirements.

If a positive integer $k$ satisfies the problem's requirements, then for the positive integer $k+20$, as long as the twenty new numbers added to $X$ are assigned according to the method described above, the maximum and minimum of these twenty numbers are assigned to $B$, and the remaining eighteen numbers to $A$, then the $k+20$ positive integers also meet the problem's requirements.

Therefore, when the positive integer $t \geqslant 8$, and $k=20 t+4$, the set $X$ has a decomposition that meets the problem's conditions.

In summary, the required positive integers are $k=20 t-1(t \in \mathbb{N})$ and $k=20 t+4$ (positive integer $t \geqslant 8$).