1. (Australia 3) The integer 9 can be expressed as the sum of two consecutive integers
$$
9=4+5=2+3+4 \text {. }
$$

Is there a positive integer that can be expressed as the sum of 1990 consecutive integers and can also be expressed in exactly 1990 ways as the sum of consecutive integers?
Let $N$ be the sum of 1990 consecutive positive integers $m$, $m+1, \cdots, m+1989$, then
$$
N=995 (2 m+1989) \text{. }
$$
Suppose $N$ can be expressed as the sum of at least two consecutive positive integers in exactly 1990 ways, i.e., there are exactly 1990 pairs of positive integers $n, k \quad(k \geqslant 2)$, such that
$$
N=\sum_{i=0}^{k-1}(n+i)=k\left(n+\frac{k-1}{2}\right) .
$$

Thus,
$$
2 N=k(2 n+k-1), \quad k \geqslant 2 .
$$

Since $k$ and $2 n+k-1$ have opposite parity, and $N$ is odd, one of them must be odd, and the other must be an odd number multiplied by 2. Noting that $k<2 n+k-1$, from (1) and (2), we can see that the number of divisors of $N$ greater than 1 is equal to the number of divisors of $2N$ greater than 1 and less than $\sqrt{2 N}$. Therefore, $2N$ has exactly 1990 divisors greater than 1 and less than $\sqrt{2 N}$, hence $2N$ has exactly 1991 divisors less than $\sqrt{2 N}$.

Since $2$, $5$, and $199$ are prime factors of $2N$, we can express $2N$ as
$$
2 N^{T}=2 \cdot 5^{a} \cdot 199^{a_{2}} p_{3}^{a^{3}} \cdots p_{5}^{\alpha},
$$

where $\alpha_{1}, \alpha_{2}$ are positive integers, $\alpha_{3}, \cdots, \sigma_{2}$ are non-negative integers, and $p_3, \cdots, p_5$ are primes different from 2, 5, and 199. This implies that $2N$ has
$$
(1+1)\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{5}+1\right)
$$

divisors, so we have $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{0}+1\right)$
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(c_{\mathrm{s}}+1\right)=1991 .
$$
Since 1991 is the product of 11 and 181, both of which are prime, we must have $\alpha_{\mathrm{s}}=\cdots=\alpha_{\mathrm{s}}=0 ; \alpha_{1}=10, \alpha_{2}=180$, or $\alpha_{1}=$ 180, $\alpha_{2}=10$. Therefore, the positive integers that satisfy the proposition are only two:
$$
N=5^{1 \times 0} 199^{10} N=5^{10} 199^{20} .
$$