$15 \cdot 21$ Can 2 be written in the form
$$
2=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\cdots+\frac{1}{n_{1974}} .
$$

where $n_{1}, n_{2}, \cdots, n_{1974}$ are distinct natural numbers.
(Kiev Mathematical Olympiad, 1974)
[Solution] First, 1 can be written as the sum of three fractions with numerator 1.
$$
1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \text {. }
$$

Thus, $\quad 2=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$.
Notice the identity
$$
\frac{1}{6 k}=\frac{1}{12 k}+\frac{1}{20 k}+\frac{1}{30 k} .
$$

Taking $k=1=5^{0}$, then
$$
\frac{1}{6}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}
$$

Substituting into (1) gives $2=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}$.
Thus, 2 is expressed as the sum of 6 fractions.
Taking $k=5^{1}=5$, then $\frac{1}{30}=\frac{1}{60}+\frac{1}{100}+\frac{1}{150}$,

Substituting into (3) gives
$$
2=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{60}+\frac{1}{100}+\frac{1}{150},
$$

Thus, 2 is expressed as the sum of 8 fractions.
Taking $k=5^{2}$ and substituting into (2) gives the decomposition of $\frac{1}{150}$, substituting into (4) thus results in 2 being expressed as the sum of 10 fractions.

Further taking $k=5^{3}, 5^{4}, \cdots, 5^{984}$, each time adding two new distinct terms to (1), after 985 such substitutions, 2 is expressed as the sum of
$$
985 \cdot 2+4=1974
$$

fractions with numerator 1.