4. There are 100 points $P_{1}, P_{2}, \cdots, P_{100}$ on a circle (where $P_{100}$ is adjacent to $P_{1}$). There are 5 colors, and each of the points $P_{1}, P_{2}, \cdots, P_{100}$ is to be colored with one of the 5 colors, with each color used at least once. If for any such coloring, there always exist $t$ consecutive points among $P_{1}, P_{2}, \cdots, P_{100}$ that contain at least 3 different colors, find the minimum value of $t$.
Let $P_{25}, P_{50}, P_{75}, P_{100}$ be colored with colors $1, 2, 3, 4$ respectively, and the rest of the points be colored with color 5. In this case, any 25 consecutive points among $P_{1}, P_{2}, \cdots, P_{100}$ contain only two colors, which does not meet the requirement, thus $t$ must be at least 26. Assume there is a coloring method such that any 26 consecutive points contain at most two colors. For this coloring method, select as many consecutive points as possible from $P_{1}, P_{2}, \cdots, P_{100}$, ensuring that these points do not contain all 5 colors, i.e., they contain exactly 4 colors (otherwise, an adjacent point can be added to these consecutive points without containing all 5 colors). Assume the selected points are $P_{k+1}, P_{k+2}, \cdots, P_{100}$, and these $100-k$ points do not contain color 1. By the maximality, $P_{1}$ and $P_{k}$ are both colored with color 1. For $i=1,2,3,4,5$, let $x_{i}$ be the maximum index of points colored with color $i$, then $x_{1}=k$. By symmetry, assume $x_{2} < x_{3} < x_{4} < x_{5}$. For each $i=1,2,3,4,5$, we prove that $x_{i}-x_{i-1} \geqslant 25$.

Otherwise, if $x_{i}-x_{i-1} \leqslant 24$, consider 26 consecutive points $P_{x_{i-1}}, P_{x_{i-1}+1}, \cdots, P_{x_{i-1}+25}$ (indices are understood modulo 100, the same applies below). Among these points, $P_{x_{i-1}}$ is colored with color $i-1$, $P_{x_{i}}$ is colored with color $i$, and $P_{x_{i}+1}$ is colored with a color $j$ different from $i-1$ and $i$ (when $i=2,3,4$, it must be that $i+1 \leqslant j \leqslant 5$; when $i=5$, since $P_{x_{5}+1}=P_{101}=P_{1}$, thus $j=1 \notin \{4,5\}$), meaning these points contain at least 3 colors, which contradicts the assumption.

Thus, $x_{5}=x_{1}+\sum_{i=2}^{5}(x_{i}-x_{i-1}) \geqslant 1+4 \times 25=101>100$, which contradicts $x_{5}=100$. The above shows that for any coloring method that meets the requirement, $P_{1}, P_{2}, \cdots, P_{100}$ always contain 26 consecutive points with at least 3 colors, so the minimum value of $t$ is 26.