Example 3 Does there exist a positive integer $n$ that satisfies the following two conditions simultaneously:
(1) $n$ can be decomposed into the sum of 1990 consecutive positive integers.
(2) $n$ has exactly 1990 ways to be decomposed into the sum of several (at least two) consecutive positive integers.
(31st IMO Shortlist)
Solution: Suppose $n$ is a natural number that meets the conditions. According to Theorem 3, condition (1) is equivalent to:
(1') $1990+\frac{2 n}{1990}$ is odd, i.e., $\frac{n}{5 \times 199}$ is odd, and $2 n \geqslant 1990 \times 1991$.
By Theorem 2, condition (2) is equivalent to:
(2) $n$ has exactly 1990 distinct odd divisors greater than 1.

From (1'), we can assume $n=5^{a} 199^{b} p_{1}^{t_{1}} p_{2}^{t_{2}} \cdots p_{r_{r}}^{t^{\prime}}$, where $p_{1}, \cdots, p_{r}$ are odd primes, $a, b, t_{1}, \cdots, t_{r}$ are positive integers, and $r \geqslant 0$.
From $\left(2^{\prime}\right)$, we have
$$
\begin{array}{l}
(a+1)(b+1)\left(t_{1}+1\right)\left(t_{2}+1\right) \cdots\left(t_{r}+1\right) \\
=1990+1=1991 .
\end{array}
$$

Noting that 1991 has only one unique factorization:
$$
1991=181 \times 11 \text {, }
$$

so, $r=0$. Thus, (6) becomes
$(a+1)(b+1)=1991=181 \times 11$.
Solving this, we get $(a, b)=(180,10),(10,180)$.
Therefore, there are only two natural numbers $n$ that meet the conditions, which are
$$
n=5^{180} \times 199^{10}, n=5^{10} \times 199^{180} \text {. }
$$