Example 3 Let the function $y=10 \tan \left[(2 k-1) \frac{x}{5}\right] \quad\left(k \in \mathbf{Z}^{+}\right)$, when $x$ varies between any two consecutive integers (including the integers themselves), it loses its meaning at least twice, find the minimum positive integer value of $k$.

When $x$ varies between any two consecutive integers (including the integers themselves), the function $y=10 \tan \left[(2 k-1) \frac{x}{5}\right]$ loses its meaning at least twice, find the minimum positive integer value of $k$.
Solve According to the problem, the smallest positive period should satisfy the following conditions:
$T \leqslant 1$, that is, $\frac{\pi}{\frac{1}{5}(2 k-1)} \leqslant 1$,
which means $k \geqslant \frac{5 \pi+1}{2}$, and since $k \in \mathbf{Z}^{+}$. Therefore, $k \geqslant 9$ is sufficient.
The solution is wrong! Where is the mistake?
Analysis This solution ignores that the function $y=10 \tan \left[(2 k-1) \frac{x}{5}\right]$ passes through the origin and is an odd function. To satisfy that $x$ changes at least twice without meaning in any two consecutive integers (including the integers themselves), it must first satisfy that $x \in[0,1]$ loses meaning twice. The correct solution is given below.
Correct solution According to the problem, the smallest positive period $T$ satisfies
$$
\frac{3}{2} T \leqslant 1 \text {, }
$$

then we have
$$
\frac{3}{2} \cdot \frac{\pi}{\frac{2 k-1}{5}} \leqslant 1,
$$

solving for $k \geqslant 13$, hence the smallest positive integer value of $k$ is 13.
Note Correct use of properties and accurate understanding of the problem are key to solving it correctly.