Let $t(A)$ denote the sum of elements of a nonempty set $A$ of integers, and define $t(\emptyset)=0$. Find a set $X$ of positive integers such that for every integers $k$ there is a unique ordered pair of disjoint subsets $(A_{k},B_{k})$ of $X$ such that $t(A_{k})-t(B_{k}) = k$.
1. **Define the Set \( X \)**:
   Let \( X = \{3^0, 3^1, 3^2, \ldots\} = \{1, 3, 9, 27, \ldots\} \). This is the set of powers of 3.

2. **Base Case**:
   For \( k = 1 \):
   - \( A_1 = \{1\} \)
   - \( B_1 = \emptyset \)
   - \( t(A_1) - t(B_1) = 1 - 0 = 1 \)

   For \( k = 2 \):
   - \( A_2 = \{3\} \)
   - \( B_2 = \{1\} \)
   - \( t(A_2) - t(B_2) = 3 - 1 = 2 \)

3. **Inductive Step**:
   Assume that for all integers \( k \) such that \( 1 \leq k \leq \frac{3^m - 1}{2} \), there exist unique disjoint subsets \( (A_k, B_k) \) of \( X \) such that \( t(A_k) - t(B_k) = k \).

4. **Extend to \( k \) in the range \( \frac{3^m - 1}{2} < k \leq 3^m \)**:
   Let \( k \) be a positive integer such that \( \frac{3^m - 1}{2} < k \leq 3^m \). Define \( l = 3^m - k \). By the inductive hypothesis, there exist unique disjoint subsets \( (A_l, B_l) \) such that \( t(A_l) - t(B_l) = l \).

   Define:
   - \( A_k = B_l \cup \{3^m\} \)
   - \( B_k = A_l \)

   Then:
   \[
   t(A_k) - t(B_k) = t(B_l \cup \{3^m\}) - t(A_l) = (t(B_l) + 3^m) - t(A_l) = 3^m - l = k
   \]

5. **Uniqueness**:
   Suppose there exist other sets \( C_k \) and \( D_k \) such that \( t(C_k) - t(D_k) = k \). We need to show that \( 3^m \) belongs to \( C_k \).

   Since \( t(C_k) \geq k > \frac{3^m - 1}{2} \), there must be a power of 3 greater than \( 3^{m-1} \) in \( C_k \). The largest such power is \( 3^m \).

   Also, \( t(D_k) \leq \frac{3^m - 1}{2} \), so:
   \[
   k \geq 3^m - \frac{3^m - 1}{2} = \frac{3^m + 1}{2} > 3^{m-1}
   \]
   This implies \( 3^m \) must be in \( C_k \).

   Let \( C_k = C'_k \cup \{3^m\} \). Then:
   \[
   t(C'_k) - t(D_k) = 3^m - k = l
   \]
   By the uniqueness of \( (A_l, B_l) \), we have \( A_l = D_k \) and \( B_l = C'_k \). Thus:
   \[
   A_k = C_k \quad \text{and} \quad B_k = D_k
   \]

6. **Extend to \( k \) in the range \( 3^m < k \leq \frac{3^{m+1} - 1}{2} \)**:
   Let \( k \) be such that \( 3^m < k \leq \frac{3^{m+1} - 1}{2} \). Define \( l = k - 3^m \). By the uniqueness of \( (A_l, B_l) \), we can find unique \( A_k \) and \( B_k \) following the same steps as above.

Thus, by induction, the set \( X = \{3^0, 3^1, 3^2, \ldots\} \) satisfies the required conditions.

The final answer is \( \boxed{ X = \{3^0, 3^1, 3^2, \ldots\} } \)