Determine for which positive integers $k$, the set \[X=\{1990, 1990+1, 1990+2, \cdots, 1990+k \}\] 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 $B$.
To determine for which positive integers \( k \), the set 
\[ X = \{1990, 1990+1, 1990+2, \ldots, 1990+k \} \]
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 \( B \), we proceed as follows:

1. **Calculate the sum of the set \( X \):**
   \[
   S(X) = \sum_{i=0}^{k} (1990 + i) = (k+1) \cdot 1990 + \sum_{i=0}^{k} i
   \]
   \[
   S(X) = 1990(k+1) + \frac{k(k+1)}{2}
   \]

2. **Check the condition for \( S(X) \) to be even:**
   For \( S(X) \) to be partitioned into two subsets with equal sum, \( S(X) \) must be even.
   \[
   S(X) = 1990(k+1) + \frac{k(k+1)}{2}
   \]
   \[
   S(X) = (k+1) \left( 1990 + \frac{k}{2} \right)
   \]
   Since \( 1990 \) is even, \( k+1 \) must be even for \( S(X) \) to be even. Therefore, \( k \) must be odd.

3. **Check divisibility conditions:**
   \[
   \frac{k(k+1)}{2} \text{ must be an integer, and } 2 \mid \frac{k(k+1)}{2} \implies 4 \mid k(k+1)
   \]
   Since \( k \) and \( k+1 \) are consecutive integers, one of them must be divisible by 4. Therefore, \( k \equiv 0 \pmod{4} \) or \( k \equiv 3 \pmod{4} \).

4. **Case 1: \( k = 4m \):**
   \[
   X = \{1990, 1990+1, \ldots, 1990+4m\}
   \]
   The number of elements in \( X \) is \( 4m+1 \), which is odd. We need to partition \( X \) into two subsets \( A \) and \( B \) with equal sums. This is not possible because the number of elements is odd, and we cannot split them into two equal-sized subsets.

5. **Case 2: \( k = 4m+3 \):**
   \[
   X = \{1990, 1990+1, \ldots, 1990+4m+3\}
   \]
   The number of elements in \( X \) is \( 4m+4 \), which is even. We can partition \( X \) into two subsets \( A \) and \( B \) with equal sums.

6. **Verification for \( k = 4m+3 \):**
   Let \( k = 4m+3 \). Then,
   \[
   S(X) = 1990(4m+4) + \frac{(4m+3)(4m+4)}{2}
   \]
   \[
   S(X) = 1990(4m+4) + \frac{(4m+3)(4m+4)}{2}
   \]
   \[
   S(X) = 1990(4m+4) + 2m(4m+4) + 2m + 6
   \]
   \[
   S(X) = 1990(4m+4) + 8m^2 + 10m + 6
   \]
   Since \( S(X) \) is even, we can partition \( X \) into two subsets \( A \) and \( B \) with equal sums.

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