Xinxin from Tao Nan School (Singapore) sent us the following solution:

Since in any subset of 4, all four students has exchanged cards with one another, then if I randomly pick four students, say index numbers 1, 4, 11, and 29, all of them has exchanged cards with one another.

How about the combination

1, 2, 3, 4? Of course they have exchanges cards with one another. 1, 5, 6, 7? Yes of course, they have.

1, 8, 9, 10? Sure they did!

1, 11, 12, 13? Yes!

1, 14, 15, 16? Yes!

1, 17, 18, 19? Yes too!

1, 20, 21, 22?

1, 23, 24, 25?

1, 26, 27, 28?

1, 29, 30, 2?

Now comes the conclusion: student number 1 has exchanged cards with everyone else in the class.

How about student number 2? Using the same method, you can find out that he or she has exchanged cards with everyone else as well.

And so did number 3, 4, and everyone else in the class.

The truth is, everyone (30 students) had exchanged cards with everyone else.

Further question: What if, in each subset of 4 students, at least one (but not necessarily all four) had exchanged cards with the other three?