2.83 Find all natural numbers $k$ such that the set
$$
S=\{1994,1997,2000, \cdots, 1994+3 k\}
$$

can be partitioned into the union of two disjoint sets $A$ and $B$, and the sum of all numbers in $A$ is 9 times the sum of all numbers in $B$.
[Solution] Clearly, $|S|=k+1$, and the sum of the $k+1$ natural numbers in $S$ is
$$
\begin{aligned}
M= & \sum_{i=0}^{k}(1994+3 j)=1994(k+1)+\frac{3}{2} k(k+1) \\
& =1990(k+1)+\frac{1}{2}(k+1)(3 k+8) .
\end{aligned}
$$

Let sets $A$ and $B$ satisfy the requirements of the problem, and let the sum of all numbers in $B$ be $x$. Thus, the sum of all numbers in $A$ is $9x$. Therefore, we have
$$
\begin{aligned}
10 x & =1990(k+1)+\frac{1}{2}(k+1)(3 k+8), \\
x & =199(k+1)+\frac{1}{20}(k+1)(3 k+8) .
\end{aligned}
$$

Since $x$ is an integer, it must be that $20 \mid (k+1)(3 k+8)$. This is equivalent to
$$
4 \mid (k+1)(3 k+8), 5 \mid (k+1)(3 k+8) .
$$

Since $(k+1)(3 k+8)=5(k+1)+3(k+1)^{2}$, the second condition in (2) is equivalent to $5 \mid (k+1)$, i.e., $k=5 m-1, m \in \mathbb{N}$. Substituting this into the first condition in (2), we get $4 \mid m(15 m+5)$. This can be simplified to $4 \mid m(3 m+1)$. Therefore, we must have $m=4 t$ or $m=4 t+1, t \in \mathbb{N}$.
(1) When $m=4 t$, $k=20 t-1, |S|=20 t$, i.e., $S$ contains $20 t$ natural numbers. Arrange the numbers in $S$ in ascending order, and divide every 20 numbers into a group starting from 1994. This results in exactly $t$ groups. The smallest and largest numbers in each group are assigned to set $B$, while the remaining 18 numbers are assigned to set $A$. It is easy to see that the sets $A$ and $B$ thus obtained satisfy the requirements of the problem.
(2) When $m=4 t+1$, $k=20 t+4, t \in \mathbb{N}$. Then from (1), we have
$$
\begin{aligned}
x & =199(20 t+5)+\frac{1}{20}(20 t+5)(60 t+20) \\
& =5(4 t+1)(3 t+200) \\
& \equiv 2(t+1) \cdot 2 \equiv t+1 \pmod{3} .
\end{aligned}
$$

Since each number in $S$ leaves a remainder of 2 when divided by 3, from (3) we also have
$$
2|B| \equiv x \equiv t+1 \pmod{3} .
$$

This implies
$$
|B|=4|B|=2 t+2 \pmod{3} .
$$

When $t=1$, $x=5075, |B| \equiv 1 \pmod{3}$. However, any number in $S$ is less than $x$ and the sum of any 4 numbers is greater than $x$, so $k=24$ does not satisfy the requirements.

When $t=2$, $x=9270, |B| \equiv 0 \pmod{3}$. However, the sum of any 3 numbers in $S$ is less than 9270 and the sum of any 6 numbers is greater than 9270, so $k=44$ does not satisfy the requirements.

Continuing in this manner, when $t=7$, $x=1990 \times 13 + 6175 = 1990 \times 16 + 205, |B| \equiv 1 \pmod{3}$. It is easy to see that the sum of the largest 13 numbers in $S$
$$
1990 \times 13 + (436 + 433 + \cdots + 400) = 1990 \times 13 + 5434
$$

is less than $x$ and the sum of the smallest 16 numbers
$$
1990 \times 16 + (4 + 7 + \cdots + 49) = 1990 \times 16 + 424
$$

is greater than $x$. Therefore, $k=144$ does not satisfy the requirements.
When $t=8$, $x=1990 \times 18 + 1140, |B| \equiv 0 \pmod{3}$.

Let
$$
B=\{1994,1997, \cdots, 2039,2210,2486\},
$$

then $|B|=18$ and the sum of these 18 numbers is
$$
\begin{aligned}
& 1990 \times 18 + (4 + 7 + \cdots + 49 + 220 + 496) \\
= & 1990 \times 18 + 1140 = x .
\end{aligned}
$$

Let $A=S-B$, then $A$ and $B$ satisfy the requirements of the problem, i.e., $t=8$ satisfies the requirements.

From (1), if a natural number $k$ satisfies the requirements of the problem, then for $k+20$, by assigning the last 20 numbers to $A$ and $B$ as in (1), it is clear that $k+20$ also satisfies the requirements. Therefore, the natural numbers $k=20 t+4 (t \in \mathbb{N}$ and $t \geqslant 8)$ all satisfy the requirements of the problem.

In summary, the required natural numbers consist of two families: $k=20 t-1, t=1,2, \cdots$ and $k=20 t+4, t=8,9, \cdots$.