Four, (50 points) 11 interest classes, several students participate (can participate repeatedly), and each interest class has the same number of students (full, unknown number). It is known that any nine interest classes include all students, while any eight interest classes do not include all students. Find the minimum total number of students.
Let the set of students in 11 interest classes be $A_{1}$, $A_{2}, \cdots, A_{11}$.
By the problem, we know $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{11}\right|=x$.
Let $T=A_{1} \cup A_{2} \cup \cdots \cup A_{11}$.
By the problem, we know the union of any nine sets is $T$, and the union of any eight sets is a proper subset of $T$.

Construct a table, if student $a_{i} \in A_{j}$, then fill the cell in the $i$-th row and $j$-th column with 1, otherwise fill it with 0.

By the condition, we know that the union of any eight sets is not $T$, i.e., for any eight columns, there must be a row where the intersecting cells are all 0, which we call a "zero row".

Furthermore, by the condition that the union of any nine sets is $T$, any two zero rows cannot be the same row.
Thus, we form a one-to-one mapping from every 8 columns to one row.
Therefore, $\mathrm{C}_{11}^{8} \leqslant|T|$.
On the other hand, each row has 3 zeros, and the columns of zeros in any two rows are not all the same, with a total of $\mathrm{C}_{11}^{3}$ ways to fill. And $\mathrm{C}_{11}^{3}=\mathrm{C}_{11}^{8}$, so each row is filled in exactly one way. At this point, the conditions of the problem are exactly satisfied.
In summary, the minimum number of students is $\mathrm{C}_{11}^{8}=165$.