C(23,12) < 2006000

(a) If there is some student whose list is a good list, we're done. Otherwise, there exist students $x, y$ so that they do not agree in any colleges. Let $X, Y$ be the sets of colleges of $x, y$, respectively. Now, split $X, Y$ into sets of size $6$, say $X = A \cup C, Y = B \cup D.$ We claim that one of the lists $X, Y, A \cup B, A \cup D, B \cup C, C \cup D$ is good. Assume the contrary, for contradiction. Suppose the contrary, for contradiction. Then, take sets $S_1, S_2, S_3, S_4$ so that they are disjoint with $A \cup B, B \cup C, C \cup D, D \cup A$ respectively. Apply the condition of the problem on $X, Y, S_1, S_2, S_3, S_4$ to obtain $v \in X$ and $w \in Y$ with each of the $S_i$'s containing $x$ or $w.$ However, this is obviously absurd. $\square$ (b) As above pointed out, observe that $\binom{23}{12} < 2006000.$ In view of this, first take $\binom{23}{12}$ and let there lists be all $12-$element subsets of $[23].$ Now, simply add $200600 - \binom{23}{12}$ other students with their lists also arbitrary subsets of $[23].$ It's easy to see that this construction prevents any $11-$element subset of colleges from being good. We therefore must show that the condition of the problem is true under this construction. Consider some six subsets $A, B, C, D, E, F \subset [23]$ with size $12.$ Notice that any two of them have a common element, and so if some four of them had a common element we would be done. Indeed, if $A, B, C, D$ had the common element $e$, e.g., then we can take the two-element subset containing $e$ and and arbitrary element in $E \cap F.$ Let's now show the existence of some four of $A, B, C, D, E, F$ which share a common element. Indeed, simply observe that $A, B, C, D, E, F$ collectively contain $12 \cdot 6 = 72$ numbers. Since $72 > 3 \cdot 23$, some $1 \le i \le 23$ is in four of $A, B, C, D, E, F.$ $\square$