Determine for which positive integers $ k$ the set \[ X \equal{} \{1990, 1990 \plus{} 1, 1990 \plus{} 2, \ldots, 1990 \plus{} k\}\] can be partitioned into two disjoint subsets $ A$ and $ B$ such that the sum of the elements of $ A$ is equal to the sum of the elements of $ B.$
1. **Sum of the Elements in Set \( X \)**:
   The set \( X \) is given by:
   \[
   X = \{1990, 1990 + 1, 1990 + 2, \ldots, 1990 + k\}
   \]
   The sum of the elements in \( X \) can be calculated as follows:
   \[
   S = \sum_{i=0}^{k} (1990 + i) = 1990(k + 1) + \sum_{i=0}^{k} i
   \]
   The sum of the first \( k \) integers is:
   \[
   \sum_{i=0}^{k} i = \frac{k(k + 1)}{2}
   \]
   Therefore, the total sum \( S \) is:
   \[
   S = 1990(k + 1) + \frac{k(k + 1)}{2}
   \]

2. **Condition for Partitioning**:
   For the set \( X \) to be partitioned into two subsets \( A \) and \( B \) with equal sums, \( S \) must be even. Thus, we need to check when \( S \) is even:
   \[
   S = 1990(k + 1) + \frac{k(k + 1)}{2}
   \]
   Simplifying \( S \):
   \[
   S = (1990 + \frac{k}{2})(k + 1)
   \]
   For \( S \) to be even, \( (1990 + \frac{k}{2})(k + 1) \) must be even. Since \( 1990 \) is even, \( 1990 + \frac{k}{2} \) is even if \( \frac{k}{2} \) is an integer, i.e., \( k \) must be even.

3. **Further Analysis**:
   We need to check the conditions under which \( S \) is even:
   - If \( k \equiv 0 \pmod{4} \), then \( k \) is divisible by 4.
   - If \( k \equiv 3 \pmod{4} \), then \( k + 1 \equiv 0 \pmod{4} \), making \( k + 1 \) divisible by 4.

4. **Verification for \( k \equiv 3 \pmod{4} \)**:
   Let \( k = 4m + 3 \) for some integer \( m \). Then:
   \[
   S = 1990(4m + 4) + \frac{(4m + 3)(4m + 4)}{2}
   \]
   Simplifying:
   \[
   S = 1990(4m + 4) + \frac{(4m + 3)(4m + 4)}{2} = 1990(4m + 4) + 2(4m + 3)(m + 1)
   \]
   Since \( 1990(4m + 4) \) is even and \( 2(4m + 3)(m + 1) \) is even, \( S \) is even.

5. **Verification for \( k \equiv 0 \pmod{4} \)**:
   Let \( k = 4m \) for some integer \( m \). Then:
   \[
   S = 1990(4m + 1) + \frac{4m(4m + 1)}{2}
   \]
   Simplifying:
   \[
   S = 1990(4m + 1) + 2m(4m + 1)
   \]
   Since \( 1990(4m + 1) \) is even and \( 2m(4m + 1) \) is even, \( S \) is even.

6. **Conclusion**:
   The set \( X \) can be partitioned into two subsets with equal sums if and only if \( k \equiv 0 \pmod{4} \) or \( k \equiv 3 \pmod{4} \).

The final answer is \( \boxed{ k \equiv 0 \pmod{4} } \) or \( k \equiv 3 \pmod{4} \).