1. The sum of seven different natural numbers is $2 n$. Is it true that there will definitely be three numbers among these seven numbers whose sum is greater than $n ?$
Solution. It is sufficient to provide a counterexample. One of the possible: $n=50$, seven numbers $11,12,13,14,15,16,19$. The sum of all seven numbers is 100. The sum $15+16+19=50$. Since these three numbers are the largest among the seven, any sum of three numbers does not exceed 50.

Supplement. To construct a counterexample, take an arithmetic progression of seven numbers with a difference of 1, such that its sum is the maximum possible less than $2 n$. The last number of this progression is replaced by a number that complements the sum to $2 n$. This algorithm works for such natural $n, k$ ( $k$-the first term of the progression), which can be related by the equation $n=4 k+6$. In the provided counterexample, $n=50, k=11$.

Answer. Not necessarily. A counterexample exists.