Given a set $S \subset N$ and a positive integer n, let $S\oplus \{n\} = \{s+n / s \in  S\}$. The sequence $S_k$ of sets is defined inductively as follows: $S_1 = {1}$, $S_k=(S_{k-1} \oplus \{k\})  \cup \{2k-1\}$ for $k = 2,3,4, ...$
(a) Determine $N - \cup _{k=1}^{\infty}  S_k$. 
(b) Find all $n$ for which $1994 \in S_n$.
Let's break down the problem into two parts and solve each part step-by-step.

### Part (a): Determine \( \mathbb{N} - \bigcup_{k=1}^{\infty} S_k \)

1. **Base Case:**
   - Start with \( S_1 = \{1\} \).

2. **Inductive Step:**
   - For \( k \geq 2 \), \( S_k = (S_{k-1} \oplus \{k\}) \cup \{2k-1\} \).
   - Here, \( S_{k-1} \oplus \{k\} = \{s + k \mid s \in S_{k-1}\} \).

3. **Inductive Hypothesis:**
   - Assume for \( k \geq 2 \), \( S_k = \left\{\sum_{i=0}^{m} (k-i) \mid m \in \mathbb{N} \text{ and } m \leq k-1\right\} \).

4. **Inductive Proof:**
   - For \( k = 2 \):
     \[
     S_2 = (S_1 \oplus \{2\}) \cup \{3\} = \{1 + 2\} \cup \{3\} = \{3\} \cup \{3\} = \{3\}
     \]
   - For \( k = 3 \):
     \[
     S_3 = (S_2 \oplus \{3\}) \cup \{5\} = \{3 + 3\} \cup \{5\} = \{6\} \cup \{5\} = \{5, 6\}
     \]
   - Continue this process to generalize:
     \[
     S_k = \left\{\sum_{i=0}^{m} (k-i) \mid m \in \mathbb{N} \text{ and } m \leq k-1\right\}
     \]

5. **Conclusion:**
   - A positive integer \( n \geq 2 \) is an element of \( \bigcup_{k=1}^{\infty} S_k \) if and only if \( n \) can be expressed as a sum of two or more consecutive positive integers.

6. **Set of Natural Numbers Not in \( \bigcup_{k=1}^{\infty} S_k \):**
   - The numbers that cannot be expressed as a sum of two or more consecutive positive integers are the powers of 2.
   - Therefore, \( \mathbb{N} - \bigcup_{k=1}^{\infty} S_k = \{2^m \mid m \in \mathbb{N}\} \).

### Part (b): Find all \( n \) for which \( 1994 \in S_n \)

1. **Express 1994 as a Sum of Consecutive Integers:**
   - We need to find \( n \) such that \( 1994 \) can be written as a sum of \( n \) consecutive integers.
   - Let the first integer in the sequence be \( a \). Then the sum of \( n \) consecutive integers starting from \( a \) is:
     \[
     1994 = a + (a+1) + (a+2) + \cdots + (a+n-1)
     \]
   - This can be simplified to:
     \[
     1994 = \frac{n(2a + n - 1)}{2}
     \]
   - Rearrange to find \( a \):
     \[
     3988 = n(2a + n - 1)
     \]

2. **Solve for \( n \):**
   - We need to find pairs \((n, a)\) such that \( 3988 = n(2a + n - 1) \).
   - Check for integer solutions by testing possible values of \( n \).

3. **Check Divisors of 3988:**
   - The divisors of 3988 are: 1, 2, 4, 7, 14, 28, 142, 1994, 3988.
   - Test each divisor to see if it satisfies the equation.

4. **Valid Solutions:**
   - After testing, we find that \( n = 7 \) is a valid solution:
     \[
     3988 = 7(2a + 6) \implies 3988 = 14a + 42 \implies 3946 = 14a \implies a = 282
     \]
   - Therefore, \( 1994 \in S_7 \).

The final answer is \( \boxed{ \mathbb{N} - \bigcup_{k=1}^{\infty} S_k = \{2^m \mid m \in \mathbb{N}\} } \) and \( n = 7 \) for which \( 1994 \in S_n \).