1. The sequence is given as $\frac{1}{2}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \ldots$. Which number stands at the 1997th position?
Solution. It is easily observed that the terms of the sequence are arranged in "groups" such that the sum of the numerator and denominator is the same for all numbers in that "group". Thus, the first "group" has one member (with the sum of the numerator and denominator equal to 2), the second "group" has two members (with the sum of the numerator and denominator equal to 3), the third group has three members (with the sum equal to 4), etc., the $n$-th group has $n$ members (with the sum of the numerator and denominator equal to $n+1$), i.e., this is the "group" $\frac{n}{1}, \frac{n-1}{2}, \frac{n-2}{3}, \ldots, \frac{2}{n-1}, \frac{1}{n}$. Conclusively, this "group" contains

$$
S_{n}=1+2+3+4+\ldots+n=\frac{n(n+1)}{2}
$$

members. Since we are asked to find which term of the sequence is at the 1997th position, we first need to determine in which "group" the desired term of the sequence is located. We have

$$
\frac{n(n+1)}{2} \leq 1997 \leq \frac{(n+1)(n+2)}{2}
$$

from which we find that $n=62$. Then $S_{62}=1953$, and there is room for 44 more members up to 1997, which are from the "group"

$$
\frac{63}{1}, \frac{62}{2}, \frac{61}{3}, \ldots, \frac{20}{44}, \ldots, \frac{1}{63}
$$

Therefore, the desired number is $\frac{20}{44}$.