TIN ** The integer 9 can be expressed as the sum of two consecutive integers: $9=4+5$. Moreover, 9 can also be expressed as the sum of consecutive integers in two ways: $9=4+5=2+3+4$. Is there a positive integer that can be expressed as the sum of 1990 consecutive positive integers and also has exactly 1990 ways to be expressed as the sum of at least two consecutive positive integers?
Let $N$ be the sum of 1990 consecutive positive integers $m, m+1, \cdots, m+1989$, then $N = 995(2m + 1989)$. $N$ is odd and has prime factors 5 and 199.

Furthermore, $N$ 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 $(n, k) (k \geq 2)$ such that $N = \sum_{i=0}^{k-1}(n+i) = k\left(n + \frac{k-1}{2}\right)$. Thus, $2N - k(2n + k - 1)$, $k \geq 2$. Therefore, $k \mid 2N$, $k < 2n + k - 1$, so for any divisor $a$ of $2N$, the smaller of $a$ and $\frac{2N}{a}$ can be taken as $k$, including the possibility of $k=1$. Hence, the number of $k$ (i.e., the number of pairs $(k, n)$) is $\frac{1}{2} d(2N) - 1$.

Since 5 and 199 are odd factors of $N$, $2N = 2 \cdot 5 \cdot 199 \cdot a_2 \cdot p_3^{a_3} \cdots p_s^{a_s}$, where $a_1, a_2 \in \mathbb{N}$, $\alpha_3, \cdots, \alpha_s \in \mathbb{N} \cup \{0\}$, and $p_3, p_4, \cdots, p_s$ are primes different from 2, 5, and 199. Thus, $(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_s + 1) = 1991 = 11 \times 181$. Noting that $\alpha_1 + 1 \geq 2$, $\alpha_2 + 1 \geq 2$, and 11 and 181 are primes, we have $\alpha_3 = \alpha_4 = \cdots = \alpha_s = 0$. Therefore, $\alpha_1 = 10$, $\alpha_2 = 180$ or $\alpha_1 = 180$, $\alpha_2 = 10$. Thus, we obtain only two numbers: $N = 5^{180} \cdot 199^{10}$ or $5^{10} \cdot 199^{180}$. Both of these numbers satisfy the requirements.