In the May 1989 issue of this journal, there was a problem from the Zhejiang Province High School Mathematics Competition as follows:

Arrange the positive rational numbers: $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \cdots$, then the position number of the number $\frac{1989}{1949}$ is $\qquad$.
The original answer is $(1+2+\cdots+3937)+1949$ $=7753902$.

In fact, the original sequence is formed by segments of fractions with the sum of the numerator and denominator being 2, 3, 4, ․ respectively. The fraction 1989/1949 has a sum of numerator and denominator of 3938, making it the 1949th number in the 3937th segment. Therefore, the sequence number is $(1+2+\cdots+3936)+$ $1949=7749965$.