Example 5 (IMO-31 Shortlist) The integer 9 can be expressed as the sum of two consecutive integers: $9=4+5$. At the same time, 9 can be written as the sum of consecutive natural numbers in two different ways: $9=4+5=2+3+4$.
Is there a natural number $N$ such that:
(1) It is the sum of 1990 consecutive integers;
(2) It can be expressed as the sum of consecutive integers in exactly 1990 different ways.
If condition (1) is satisfied, then
$$
\begin{aligned}
N & =n+(n+1)+(n+2)+\cdots+(n+1989) \\
& =\frac{1}{2} \cdot 1990(2 n+1989) \\
& =995(2 n+1989) .
\end{aligned}
$$

By (1), $N$ must be an odd number.
Suppose $N$ can be written as the sum of $k+1$ (where $k$ is a natural number) consecutive natural numbers, and the smallest of these $k+1$ numbers is $m$, then
$$
\begin{aligned}
N & =m+(m+1)+\cdots+(m+k) \\
& =\frac{1}{2}(k+1)(2 m+k),
\end{aligned}
$$

which means $2 N=(k+1)(2 m+k)$.
By (1), we have
$$
1990(2 n+1989)=(k+1)(2 m+k) \text {. }
$$

Therefore, the number of pairs $(m, k)$ that satisfy the above equation is the number of ways $N$ can be written as the sum of $k+1$ consecutive natural numbers.

By (2) and the fact that $m$ is a natural number, we have
$$
k+1 \mid 2 N, \quad k+1<k+2 m \text {. }
$$

So $k+1<\sqrt{2 N}$.
Conversely, if there is a divisor of $2 N$ that is less than $\sqrt{2 N}$, let this divisor be $k+1$, then
$$
2 N=(k+1) c \text {. }
$$

where $k+1<c$.
$c$ can be written as $2 m+k$, where $m \in \mathbf{N}$.
Since $N$ is odd, $k+1$ and $c$ cannot both be even. Therefore, when $k$ is odd, $c$ is odd; when $k$ is even, $c$ is even. Therefore, $c$ can be written as $2 m+k$.
Thus, $2 N=(k+1)(2 m+k)$.
This shows that when $N$ is odd, if $N$ can be written as the sum of more than one consecutive natural numbers in $l$ ways, then $l$ is the number of divisors of $2 N$ that are greater than 1 and less than $\sqrt{2 N}$.
By (1) $N=995(2 n+1989)$, we get
$$
2 N=1990(2 n+1989)=2 \cdot 5 \cdot 199(2 n+1989)=2^{1} \cdot 5^{a} \cdot 199^{b} p_{1}^{b_{1}} p_{2}^{b_{2}} \cdots p_{k^{b}}^{b_{k}} \text {. }
$$

Since $N$ has 1990 different ways of division, including the case of 1 natural number, the number of divisors of $2 N$ is
$$
(1+1)(a+1)(b+1)\left(b_{1}+1\right) \cdot \cdots \cdot\left(b_{k}+1\right)=2 \cdot 1991=2 \cdot 11 \cdot 181 .
$$

where $a, b \in \mathbf{N}, b_{i} \in \mathbf{N} \bigcup\{0\}, i=1,2, \cdots, k$.
Thus, $b_{1}=b_{2}=\cdots=b_{k}=0$, so
$(a+1)(b+1)=11 \cdot 181$, then
$\left\{\begin{array}{l}a=10, \\ b=180,\end{array}\right.$ or $\left\{\begin{array}{l}a=180, \\ b=10 .\end{array}\right.$
Thus, $N=5^{10} \cdot 199^{180}$ or $N=5^{180} \cdot 199^{10}$.