The least common multiple of two numbers is 19 more than their greatest common divisor. What could these two numbers be?
Let's find the positive integer solutions. Denote the least common multiple of the two numbers we are looking for by $t$, and their greatest common divisor by $d$. The least common multiple is divisible by each of the numbers, and thus by their greatest common divisor as well, so the difference $t-d$, which according to the problem is 19, is divisible by $d$. Since 19 is a prime number, its (positive) divisors are only 1 and 19, so only these can be considered for $d$.

If $d=1$, then $t=1+19=20$ is the least common multiple of the two numbers we are looking for, and the numbers are relatively prime, so their least common multiple is their product. The factorizations of 20 into two relatively prime factors (ignoring the order of the factors) are $1 \cdot 20$ and $4 \cdot 5$, so 1,20 and 4,5 are two suitable pairs of numbers.

If $d=19$, then $t=38$, so the two numbers are multiples of 19 and divisors of 38, hence 19 or 38. The two numbers cannot be equal, because then $t-d=0$ would be, but the pair 19, 38 satisfies the conditions of the problem. - Thus, three pairs of positive integers satisfy the requirements of the problem.

If we also consider negative integers, then the pairs of numbers that meet the conditions are those that arise from the above by changing the sign of one or both numbers, since a number and its negative have the same multiples and divisors. Thus, the least common multiple and greatest common divisor of two numbers are the same as those of the absolute values of the two numbers.

Note. Many contestants only wrote down one or two solutions, or all three, but without any justification. In addition to these, the problem could have had any number of other solutions. It was the essential mathematical idea that the realization that the above three pairs of numbers give all the solutions was required.