1. Four teams participated in a hockey tournament, with each team playing exactly one match against each of the others. The number of goals scored in each match divides the total number of goals scored in the tournament, and no two matches had the same number of goals. What is the minimum number of goals that could have been scored in the tournament?
1. If each team plays one match against each other, each team will play a total of three matches in the tournament, and the total number of matches will be $\frac{1}{2} \cdot 4 \cdot 3=6$. Therefore, we need to find six different natural numbers (zero does not divide any number) with the smallest possible sum such that this sum is divisible by each of the six addends. The smallest sum of six different natural numbers is $1+2+3+4+5+6=21$, but this is not divisible by, for example, two or six. The next possibility is to replace the number 6 with 7, making the sum 22. However, 22 is not divisible by, for example, three. The sum 23 cannot work because 23 is a prime number, divisible only by two natural numbers. Finally, the number 24 is the sum of the numbers $1,2,3,4,6$, and 8, and 24 is divisible by each of the numbers $1,2,3,4,6,8$. Therefore, 24 goals could have been scored in the tournament, but not fewer.

For determining the number of 6 matches and estimating that $1+2+3+4+5+6$ is at most equal to the number of goals, award 1 point; for excluding the sums 21, 22, and 23, award 2 points; for determining the suitable set of six numbers $1,2,3,4,6,8$ with a sum of 24, award another three points, thus 6 points for a complete solution.