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月2 * There are two four-digit numbers, A and B. The common logarithm of A is $a+\lg b$, where $a$ and $b$ are non-zero digits. The sum of the thousand's digit and the unit's digit of B is $\frac{5}{2} b$, and B is $5 b+1$ less than half of A. Find these two four-digit numbers.
Let the first number be $x$, and the second number be $y$, then $\lg x=a+\lg b=\lg \left(10^{a} \cdot b\right)$, so $x=10^{a} \cdot b$. Since the first number is a four-digit number, we have $a=3, x=10^{3} \cdot b$.

The sum of the thousands digit and the units digit of the second number is $\frac{5}{2} b$. Therefore, $\frac{5}{2} b \leqslant 18$. Since $\frac{5}{2} b$ is an integer, $b$ must be 2, 4, or 6.
If $b=2$, then $x=2000, y=\frac{1}{2} x-(5 b+1)=989$, which does not meet the requirement.
If $b=6$, then $x=6000, y=\frac{x}{2}-(5 b+1)=2969$, in this case, the sum of the thousands digit and the units digit of $y$ is $2+9=11 \neq \frac{5}{2} b=15$, so $b \neq 6$.
If $b=4$, then $x=4000, y=1979$, which meets the condition.
Therefore, the two four-digit numbers are 4000 and 1979.