Let $k$ be a positive integer with the following property: For every subset $A$ of $\{1,2,\ldots, 25\}$ with $|A|=k$, we can find distinct elements $x$ and $y$ of $A$ such that $\tfrac23\leq\tfrac xy\leq\tfrac 32$.  Find the smallest possible value of $k$.
1. **Understanding the Problem:**
   We need to find the smallest positive integer \( k \) such that for every subset \( A \) of \(\{1, 2, \ldots, 25\}\) with \(|A| = k\), there exist distinct elements \( x \) and \( y \) in \( A \) satisfying \(\frac{2}{3} \leq \frac{x}{y} \leq \frac{3}{2} \).

2. **Analyzing the Condition:**
   Consider the subset \( A = \{1, 2, 4, 7, 11, 17\} \). This subset does not satisfy the given property because for any pair \( (x, y) \) in \( A \), the ratio \(\frac{x}{y}\) does not lie within \(\left[\frac{2}{3}, \frac{3}{2}\right]\).

3. **Generalizing the Condition:**
   For a subset \( A \) with \( |A| = k \), denote the elements of \( A \) as \( 1 \leq a_1 < a_2 < \cdots < a_k \leq 25 \). The condition that no pair \( (x, y) \) satisfies \(\frac{2}{3} \leq \frac{x}{y} \leq \frac{3}{2}\) translates to:
   \[
   2a_{n+1} \geq 3a_n + 1 \quad \text{for all} \quad 1 \leq n \leq k-1
   \]
   This inequality can be rewritten as:
   \[
   a_{n+1} + 1 \geq \frac{3}{2}(a_n + 1)
   \]

4. **Finding the Smallest \( k \):**
   We need to find the smallest \( k \) such that the above inequality cannot hold for all \( 1 \leq n \leq k-1 \). Let's start with \( a_1 = 1 \) and apply the inequality iteratively:
   \[
   a_2 + 1 \geq \frac{3}{2}(a_1 + 1) \implies a_2 + 1 \geq \frac{3}{2}(1 + 1) = 3 \implies a_2 \geq 2
   \]
   \[
   a_3 + 1 \geq \frac{3}{2}(a_2 + 1) \implies a_3 + 1 \geq \frac{3}{2}(2 + 1) = 4.5 \implies a_3 \geq 4
   \]
   \[
   a_4 + 1 \geq \frac{3}{2}(a_3 + 1) \implies a_4 + 1 \geq \frac{3}{2}(4 + 1) = 7.5 \implies a_4 \geq 7
   \]
   \[
   a_5 + 1 \geq \frac{3}{2}(a_4 + 1) \implies a_5 + 1 \geq \frac{3}{2}(7 + 1) = 12 \implies a_5 \geq 11
   \]
   \[
   a_6 + 1 \geq \frac{3}{2}(a_5 + 1) \implies a_6 + 1 \geq \frac{3}{2}(11 + 1) = 18 \implies a_6 \geq 17
   \]
   \[
   a_7 + 1 \geq \frac{3}{2}(a_6 + 1) \implies a_7 + 1 \geq \frac{3}{2}(17 + 1) = 27 \implies a_7 \geq 26
   \]
   Since \( a_7 \) must be at most 25, the inequality fails for \( k = 7 \).

5. **Conclusion:**
   Therefore, the smallest possible value of \( k \) such that every subset \( A \) of \(\{1, 2, \ldots, 25\}\) with \(|A| = k\) contains distinct elements \( x \) and \( y \) satisfying \(\frac{2}{3} \leq \frac{x}{y} \leq \frac{3}{2}\) is \( k = 7 \).

The final answer is \( \boxed{7} \).