First, prove: $n_{\text {min }}>48$.
Draw a line segment $AB$ of length 5 on the plane, and construct two circles with radii of 0.5 centered at $A$ and $B$, respectively. Take 24 points in each circle. Then there are 48 points on the plane that satisfy the problem's condition (any three points must have at least two points with a distance no greater than 1).

Obviously, it is impossible to construct a circle with a radius of 1 that contains 25 of the selected points.
Therefore, $n_{\text {min }}>48$.
Next, prove: $n_{\text {min }}=49$.
If $n=49$, let $A$ be one of the points. Construct a circle $\odot A$ with a radius of 1. If all the points are within $\odot A$, then the condition of the problem is satisfied.

Otherwise, there is at least one point $B$ not in $\odot A$. Construct another circle $\odot B$ with a radius of 1. Then the distance between points $A$ and $B$ is greater than 1 (as shown in Figure 4).

Thus, for the remaining 47 points, each point $P$ forms a triplet with $A$ and $B$, and it must be true that $PA \leqslant 1$ or $PB \leqslant 1$, meaning point $P$ is either in $\odot A$ or in $\odot B$.

According to the pigeonhole principle, one of the circles must contain at least 24 of these 47 points (let's assume it is $\odot A$). Adding the center point $A$, there are at least 25 points in the circle $\odot A$ with a radius of 1 (inside or on the circumference).
Therefore, the minimum value of $n$ is 49.