1. In each cell of a $100 \times 100$ table, a natural number is written. In each row, there are at least 10 different numbers, and in any four consecutive rows, there are no more than 15 different numbers. What is the maximum number of different numbers that can be in the table?
Answer: 175

Solution. In one line, there are no less than 10 different numbers, so in the next three lines together, there appear no more than 5 new numbers. Therefore, the first four lines contain no more than 15 different numbers, and each of the following three lines adds no more than 5 new numbers, making the total number of numbers no more than $15+32 \cdot 5=175$.

Let's provide an example with 175 numbers. We will number the lines from 1 to 100. In the first line, we will place numbers from 1 to 10, and in the lines with numbers from $3k-1$ to $3k+1$, we will place numbers 1 to 5 and numbers from $5k+6$ to $5k+10$. Then, in each line, there will be 5 unique numbers and numbers from 1 to 5, i.e., exactly 10 different numbers, and in every four lines, there will be exactly 15 different numbers. Thus, in the table, there will be numbers from 1 to $5 \cdot 33 + 10 = 175$.

Remark. It can also be proven by induction that the number of different numbers in the table does not exceed 175. Specifically, it can be proven that in any $3n+1$ consecutive lines, there are no more than $5(n+2)$ different numbers. The base case $n=1$ is true by the condition. We will establish the transition from $n$ to $n+1$. Consider $3n+4$ consecutive lines. Suppose the fourth line from the bottom contains $k \geqslant 10$ different numbers. Then, in the three lowest lines, there are no more than $15-k$ different numbers. And in the remaining $3n+1$ lines, by the inductive hypothesis, there are no more than $5(n+2)$ numbers. Therefore, the total number of different numbers will be more than $5(n+2)+15-k=5(n+5)-k \leqslant 5(n+3)$.