14. (12 points) There is a sequence of numbers: $\frac{1}{1}, \frac{1}{2}, \frac{2}{2}, \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{1}{4}, \cdots$. What is the sum of the first 1996 numbers?
【Solution】Solution: There is 1 number with 1 as the denominator, the sum $S 1=1$,
There are 2 numbers with 2 as the denominator, the sum $S 2=\frac{1}{2}+\frac{2}{2}=\frac{3}{2}$,
There are 3 numbers with 3 as the denominator, the sum $S 3=\frac{1}{3}+\frac{2}{3}+\frac{3}{3}=2, \cdots$
There are $N$ numbers with $N$ as the denominator, the sum $S N=\frac{1}{N}+\frac{2}{N}+\cdots \frac{N}{N}=\frac{N(N+1)}{2N}=\frac{N+1}{2}$,
To find the sum of the first 1996 numbers, first determine what the denominator of the 1996th number is, i.e., find the smallest integer $N$ that satisfies $1+2+3+4 \cdots+N=\frac{N(N+1)}{2} \geqslant 1996$, it is easy to get $N=63,62 \times \frac{63}{2}=1953$,
There are $1996-1953=43$ numbers with 63 as the denominator, i.e., $\frac{1}{63}, \frac{2}{63}, \frac{3}{63} \cdots \frac{43}{63}$,
Then the sum of the first 1996 numbers is, $S=S 1+S 2+\cdots S 62+\frac{1}{63}+\frac{2}{63}+\cdots \frac{43}{63}$,
$=(62+1+2+3+\cdots 62) \div 2+(1+2+3 \cdots+43) \div 63,=1022.52$; Answer: The sum of its first 1996 numbers is 1022.52.