2. $42 N$ is the set of all positive integers. For a subset $S$ of $N$ and $n \in N$, define
$$
S \oplus\{n\}=\{s+n \mid s \in S\} .
$$

Additionally, define the subset $S_{k}$ as follows:
$$
S_{1}=\{1\}, S_{k}=\left\{S_{k-1} \oplus\{k\}\right\} \cup\{2 k-1\}, k=2,3,4 \cdots .
$$
(1) Find $N-\bigcup_{k=1}^{\infty} S_{k}$.
(2) Find all $k \in N$ such that $1994 \in S_{k}$.
［Solution］(1) Since
$$
\begin{array}{l} 
C_{k+1}^{2}+(k+1)= \frac{1}{2} k(k+1)+(k+1) \\
= \frac{1}{2}(k+1)(k+2) \\
=C_{k+2}^{2}, \\
C_{k+2}^{2}-C_{k}^{2}= \frac{1}{2}(k+1)(k+2)-\frac{1}{2} k(k-1)=2 k+1
\end{array}
$$

By the problem statement, we have
$$
\begin{array}{l}
S_{1}=\{1\}, \\
S_{2}=\left\{S_{1} \oplus\{2\}\right\} \cup\{3\}=\{3\}=\left\{C_{3}^{2}\right\}, \\
S_{3}=\left\{S_{2} \oplus\{3\}\right\} \cup\{5\}=\{6,5\}=\left\{C_{4}^{2}, C_{4}^{2}-C_{2}^{2}\right\},
\end{array}
$$

Assume for some positive integer $k$,
$$
S_{k}=\left\{C_{k+1}^{2}, C_{k+1}^{2}-C_{2}^{2}, C_{k+1}^{2}-C_{3}^{2}, \cdots, C_{k+1}^{2}-C_{k-1}^{2}\right\},
$$

Then by the problem statement, we have
$$
\begin{aligned}
S_{k+1}= & \left\{S_{k} \oplus\{k+1\}\right\} \cup\{2 k-1\} \\
= & \left\{C_{k+1}^{2}+k+1, C_{k+1}^{2}-C_{2}^{2}+k+1, C_{k+1}^{2}-C_{3}^{2}+k+1,\right. \\
& \left.\cdots, C_{k+1}^{2}-C_{k-1}^{2}+(k+1), 2 k+1\right\} \\
= & \left\{C_{k+2}^{2}, C_{k+2}^{2}-C_{2}^{2}, C_{k+2}^{2}-C_{3}^{2}, \cdots, C_{k+2}^{2}-C_{k-1}^{2}, C_{k+2}^{2}-\right. \\
& \left.C_{k}^{2}\right\}
\end{aligned}
$$

Therefore, for any positive integer $k$, we have
$$
S_{k}=\left\{C_{k+1}^{2}, C_{k+1}^{2}-C_{2}^{2}, C_{k+1}^{2}-C_{3}^{2}, \cdots, C_{k+1}^{2}-C_{k-1}^{2}\right\}
$$

For $j=1,2,3, \cdots, k-1$, we have
$$
\begin{array}{r}
C_{k+1}^{2}-C_{j}^{2}=\frac{1}{2} k(k+1)-\frac{1}{2} j(j-1) \\
\frac{1}{2}(k+j)(k-j+1) \in S_{k} .
\end{array}
$$

Where $C_{1}^{2}=0$, and $k+j$ and $k-j+1$ have different parities, so neither side of the above equation can be written in the form $2^{m}(m \in N)$. Conversely, if a positive integer $n$ cannot be written in the form $2^{m}(m \in N)$, then $n$ must have an odd prime factor, and thus there exist two positive integers $p, q$ with different parities, such that $p>q \geqslant 2$, and
$$
2 n=p q \text {. }
$$

Let
$$
k+j=p, k-j+1=q
$$

Then
$$
k=\frac{1}{2}(p+q-1), j=\frac{1}{2}(p-q+1)
$$

Since $p, q$ have different parities and $p \geqslant q+1$, $k, j$ are both positive integers, and
$$
k-j=q-1 \geqslant 1,1 \leqslant j \leqslant k-1 .
$$

Thus (2) can be expressed as
$$
\begin{aligned}
n & =\frac{1}{2}(k+j)(k-j+1) \\
& =C_{k+2}^{2}-C_{j}^{2} \in S_{k} .
\end{aligned}
$$

Where $C_{1}^{2}=0$.
Therefore
$$
N-\bigcup_{k=1}^{\infty} S_{k}=\left\{2^{m} \mid m \in N\right\} .
$$
(2) Since $n=1994$, we have
$$
2 n=4 \cdot 997 \text {, }
$$

Note that 997 is a prime number, so in the equation (2) $2 n=p q$,
$$
p=997, q=4
$$

From (3) we get
$$
k=500, j=497
$$

Thus
$$
1994 \in S_{500}
$$

That is, the only $k$ is: $k=500$.