Four, (20 points) Two three-digit numbers written together form a six-digit number. If this six-digit number is exactly an integer multiple of the product of the original two three-digit numbers, find this six-digit number.
Let the six-digit number be abcdef.
By the problem, we can set $\overline{a b c d e f}=k \overline{a b c} \cdot \overline{d e f}$, that is,
$\overline{a b c} \cdot 1000+\overline{d e f}=k \overline{a b c} \cdot \overline{d e f}$.
Therefore, $\overline{a b c} \mid \overline{d e f}$.
Let $\overline{\operatorname{def}}=l \overline{a b c}(l \in\{1,2, \cdots, 9\})$. Thus,
$k l \underline{a b c^{2}}=(1000+l) a b c$,
which means $k l \overline{a b c}=1000+l$.
Therefore, $l \mid 1000$.
Hence, $l=1$ or 2 or 5.
If $l=1$, then $k \overline{a b c}=1001=7 \times 11 \times 13$, it can only be
$k=7, \overline{a b c}=143$, at this time, the six-digit number
$\overline{a b c d e f}=143143$;
If $l=2$, then $\overline{a b c}<500$, and $2 k \overline{a b c}=1002$, $k \overline{a b c}=501=3 \times 167$, it can only be $k=3, \overline{a b c}=167$, at this time, the six-digit number abcdef $=167334$;

If $l=5$, then $\overline{a b c}<200$, and $5 k \overline{a b c}=1005$, $k \overline{a b c}=201=3 \times 67$, this is impossible.
In summary, the six-digit numbers sought are 143143 and 167334.