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 $\mathrm{C}_{11}^{8}=165$.