31. (9th grade) When a six-digit number, consisting of identical digits, is divided by a four-digit number, also consisting of identical digits, the quotient is 233 and there is some remainder. After discarding one digit from both the dividend and the divisor, and performing the division again, the quotient remains unchanged, but the remainder decreases by 1000. Find these numbers.
31. The sought numbers have the form $N_{1}=111111 a, N_{2}=1111 b$. We can express the condition of the problem with the following equations:

$$
\begin{gathered}
111111 a=1111 b \cdot 233+p \\
11111 a=111 b \cdot 233+p-1000
\end{gathered}
$$

By performing term-by-term subtraction, we get $100 a=233 b+1$. From this, $a=2 b+\frac{33 b+1}{100}$, $a$ is an integer when $b=3$. Answer: $N_{1}=777777, N_{2}=3333$.

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