Problem 811

$1 . \because 1990 \mid n$,
$\therefore$ Let $n=1990 k=2 \times 5 \times 199 k, k \in \mathbf{N}_{+}$.
Let $k=2^{\alpha} \cdot 5^{\beta} \cdot 199^{\gamma} \cdot q_{1}^{\gamma_{1}} \cdot q_{2}^{\gamma_{2}} \cdots q_{t^{\gamma_{1}}}^{\gamma_{1}}, \alpha, \beta, \gamma, \gamma_{i} \in \mathbf{N}(i=1,2, \cdots, t)$,
$\therefore$ We have $n=2^{a+1} \cdot 5^{\beta+1} \cdot 199^{r+1} \cdot q_{1}^{\gamma_{1}} \cdot q_{2}^{\gamma_{2}} \cdots q_{t}^{\gamma_{t}}$.
Also, $n$ has exactly 12 positive divisors,
$\therefore \quad 12=(\alpha+1+1)(\beta+1+1)(\gamma+1+1)\left(\gamma_{1}+1\right) \cdots\left(\gamma_{t}+1\right)$,
which means $2 \times 2 \times 3=(\alpha+2)(\beta+2)(\gamma+2)\left(\gamma_{1}+1\right) \cdots\left(\gamma_{t}+1\right)$,
$\therefore$ We have $\gamma_{i}=0, i-1,2, \cdots, t$,
$\alpha+2, \beta+2, \gamma+2$ have one value of 3, and the other two are 2.
$\therefore$ When $\alpha=1$, $\beta=\gamma=0$, we have $n=2^{2} \cdot 5 \cdot 199$,
When $\beta=1$, $\alpha=\gamma=0$, we have $n=2 \cdot 5^{2} \cdot 199$,
When $\gamma=1$, $\alpha=\beta=0$, we have $n=2 \cdot 5 \cdot 199^{2}$.

In summary, $n$ has 3 values, namely $2^{2} \cdot 5 \cdot 199 ; 2 \cdot 5^{2} \cdot 199 ; 2 \cdot 5 \cdot 199^{2}$.