(1) the statement is true for 2 people (and less than two doesn't count as a group). Because each must be friends with the other.

(2) if true for n $\ge$ 2 people and a new one arrives, consider separately n even or odd.

If n = 2k, then the new arrival makes a group of 2k + 1, and by the premiss of the problem, everyone in this new group must be friends with at least k + 1 of the group. So the new arrival has k + 1 friends already seated and at most k - 1 people separating them so he must have two adjacent friends he can be placed between.

If n = 2k + 1 his arrival makes 2k + 2 and he must have k + 1 friends in that group. Again they are already seated with at most k people separating them so there must be two adjacent that he can be seated between.

So, in either case the statement is true for n + 1, and is therefore true for all n $\ge$ 2.

The key to the solution is that with the arrival of a new person he has in any case a minimum of k + 1 friends among 2k or 2k + 1 people already seated.

Base Case: The statement is true for $2$ people (and less than two doesn't count as a group). Because each must be friends with the other.

Induction Hypothesis: Let the statement be true for $n$ people.

Inductive Step: If true for $n \ge 2$ people and a new one arrives, consider separately $n$ even or odd.

If $n = 2k$, then the new arrival makes a group of $2k + 1$, and by the premise of the problem, everyone in this new group must be friends with at least $k + 1$ of the group. So the new arrival has $k + 1$ friends already seated and at most $k - 1$ people separating them so he must have two adjacent friends he can be placed between.

If $n = 2k + 1$ his arrival makes $2k + 2$ and he must have $k + 1$ friends in that group. Again they are already seated with at most $k$ people separating them so there must be two adjacent that he can be seated between.

So, in either case the statement is true for $n + 1$, and is therefore true for all $n \ge 2$.

The key to the solution is that with the arrival of a new person he has in any case a minimum of $k + 1$ friends among $2k$ or $2k + 1$ people already seated.

(1) the statement is true for 2 people (and less than two doesn't count as a group). Because each must be friends with the other.

(2) if true for n people and a new one arrives, consider separately n even or odd.

If n = 2k, then the new arrival makes a group of 2k + 1, and by the premiss of the problem, everyone in this new group must be friends with at least k + 1 of the group. So the new arrival has k + 1 friends already seated and at most k - 1 people separating them so he must have two adjacent friends he can be placed between.

If n = 2k + 1 his arrival makes 2k + 2 and he must have k + 1 friends in that group. Again they are already seated with at most k people separating them so there must be two adjacent that he can be seated between.

The key to the solution is that with the arrival of a new person he has in any case a minimum of k + 1 friends among 2k or 2k + 1 people already seated.

(1) the statement is true for 2 people (and less than two doesn't count as a group). Because each must be friends with the other.

(2) if true for n $\ge$ 2 people and a new one arrives, consider separately n even or odd.

If n = 2k, then the new arrival makes a group of 2k + 1, and by the premiss of the problem, everyone in this new group must be friends with at least k + 1 of the group. So the new arrival has k + 1 friends already seated and at most k - 1 people separating them so he must have two adjacent friends he can be placed between.

If n = 2k + 1 his arrival makes 2k + 2 and he must have k + 1 friends in that group. Again they are already seated with at most k people separating them so there must be two adjacent that he can be seated between.

So, in either case the statement is true for n + 1, and is therefore true for all n $\ge$ 2.

The key to the solution is that with the arrival of a new person he has in any case a minimum of k + 1 friends among 2k or 2k + 1 people already seated.