What is the minimum $k{}$ for which among any three nonzero real numbers there are two numbers $a{}$ and $b{}$ such that either $|a-b|\leqslant k$ or $|1/a-1/b|\leqslant k$?

[i]Maxim Didin[/i]
1. We start by noting that the set of numbers \(-1, \frac{1}{2}, 2\) provides a solution, so \(k \geq \frac{3}{2}\).

2. Assume that there is a solution \(k > \frac{3}{2}\). Suppose the three numbers are \(a, b, c\) where \(c < a < b\).

3. There cannot be three positive numbers, otherwise either \(c, a \in (0, 1]\) or \(\frac{1}{a}, \frac{1}{b} \in (0, 1]\), and the solution would not be \(k > \frac{3}{2}\). Similarly, there cannot be three negative numbers.

4. Since \(a, b, c\) is a solution if and only if \(-a, -b, -c\) is a solution if and only if \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) is a solution, we can assume without loss of generality that \(c < 0 < a < b\) and \(ab \leq 1\).

5. If \(a = 1\), then \(\left|\frac{1}{a} - \frac{1}{b}\right| < \frac{3}{2}\). If \(b = 1\), then \(|b - a| < \frac{3}{2}\). Therefore, \(c < 0 < a < 1 < b\).

6. We have \(b \geq a + k\) and \(c \leq a - k\). Also, \(\frac{1}{c} \leq \frac{1}{b} - k \leq \frac{1}{a + k} - k\).

7. Since \(c < 0\) and \(a < 1 < k\), \(\frac{1}{a + k} < 1 < k\). We have \((a - k)\left(\frac{1}{a + k} - k\right) \leq c \times \frac{1}{c} = 1\).

8. This implies:
   \[
   (a - k)\left(\frac{1}{a + k} - k\right) \leq 1
   \]
   \[
   (a - k)\left(\frac{1}{a + k} - k\right) \leq 1
   \]
   \[
   (a - k)\left(\frac{1}{a + k} - k\right) \leq 1
   \]
   \[
   k^3 + ak^2 - k - ak^2 - a^2k + a \leq a + k
   \]
   \[
   k^3 - a^2k \leq 2k
   \]
   \[
   k^2 - a^2 \leq 2
   \]
   \[
   a \geq \sqrt{k^2 - 2}
   \]
   \[
   b \geq \sqrt{k^2 - 2} + k
   \]

9. From the assumption, \(1 \geq ab \geq \sqrt{k^2 - 2}(\sqrt{k^2 - 2} + k) = k^2 - 2 + k\sqrt{k^2 - 2}\).

10. This implies:
    \[
    3 - k^2 \geq k\sqrt{k^2 - 2}
    \]
    \[
    3 - k^2 \geq 0
    \]
    \[
    k^4 - 6k^2 + 9 \geq k^4 - 2k^2
    \]
    \[
    4k^2 \leq 9
    \]
    \[
    k \leq \frac{3}{2}
    \]

Thus, we have shown that \(k \leq \frac{3}{2}\), and since we already know \(k \geq \frac{3}{2}\), it follows that \(k = \frac{3}{2}\).

The final answer is \(\boxed{\frac{3}{2}}\).