1.4. [5] In the fields of a $29 \times 29$ table, each of the natural numbers $1,2, \ldots, 29$ is written 29 times. If the sum of the numbers written in the fields above the main diagonal is three times the sum of the numbers written below it, which number is written in the central field of the table? (The main diagonal connects the upper left with the lower right corner of the table.)

Answer. 15.
Solution. Above the diagonal, there are $1+2+\cdots+28=\frac{28.29}{2}=406$ numbers; there are as many numbers below the diagonal. However, the sum of the 406 largest numbers in the table (the numbers $16,17, \ldots, 29$, taken 29 times each) is equal to $29 \frac{(16+29) .14}{2}=29.45 .7$ and is exactly three times greater than the sum of the 406 smallest numbers (the numbers $1,2, \ldots, 14$ taken 29 times each), which is $29 \frac{(1+14) .14}{2}=29.15 .7$. Therefore, all the numbers on the diagonal are equal to 15.