4.26 The integer 9 can be expressed as the sum of two consecutive integers, $9=4+5$, and 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.
(31st International Mathematical Olympiad Preliminary Question, 1990)
[Solution] 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
$$
\begin{aligned}
N & =m+(m+1)+\cdots+(m+k) \\
& =\frac{1}{2}(k+1)(2 m+k), \\
2 N & =(k+1)(2 m+k) .
\end{aligned}
$$

By (1) we have
$$
1990(2 n+1989)=(k+1)(2 m+k) .
$$

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
$$
k+1<\sqrt{2 N} \text {. }
$$

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

where $k+1<c$.
$c$ can be written as $2 m+k, m \in N$.
Since $N$ is odd, $k+1$ and $c$ cannot both be even. Therefore, when $k$ is odd, $c$ is odd, and when $k$ is even, $c$ is even, so $c$ 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)
$$
\begin{aligned}
& N=995(2 n+1989), \\
2 N & =1990(2 n+1989) \\
& =2 \cdot 5 \cdot 199(2 n+1989) \\
& =2^{1} \cdot 5^{a} \cdot 199^{b} p_{1}^{b_{1}} p_{2^{2}}^{b^{2}} p_{k^{k}}^{b}
\end{aligned}
$$

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

where $\quad a, b \in N, \quad b_{i} \in N \cup\{0\}, \quad i=1,2, \cdots, k$.
Thus
$$
\begin{array}{l}
b_{1}=b_{2}=\cdots=b_{k}=0, \\
(a+1)(b+1)=11 \cdot 181,
\end{array}
$$

then
$$
\left\{\begin{array} { l } 
{ a = 1 0 , } \\
{ b = 1 8 0 , }
\end{array} \text { or } \quad \left\{\begin{array}{l}
a=180, \\
b=10 .
\end{array}\right.\right.
$$

That is, $N=5^{10} \cdot 199^{180}$ or $N=5^{180} \cdot 199^{10}$.