A 100-row and 100-column table has all 1s in its first column, and in the $k$-th row, there is an arithmetic sequence with a difference of $k$. Among the numbers located along the diagonal stretching from the bottom left to the top right of the table, which is the largest?
The bottom left corner of the table contains the number 1. The diagonal passes through the 2nd number in the row above it. This is the 2nd term of an arithmetic sequence whose first term is 1, and the common difference is 99, so the number is $1+1 \cdot 99=100$. In the next row, the diagonal passes through the 3rd number, which is $1+2 \cdot 98=199$, and so on.

In general, consider an $n \times n$ table filled in a similar manner. The mentioned diagonal will pass through the $(n-k+1)$-th position in the $k$-th row $(k=1,2,3, \ldots, n)$, which can be expressed as

$$
a_{k}=1+(n-k+1-1) k=1+n k-k^{2}=-\left(k-\frac{n}{2}\right)^{2}+\frac{n^{2}}{4}+1
$$

We need to find the largest of these numbers. The quadratic expression in $k$ has a negative leading coefficient (the coefficient of $k^{2}$), so the parabola opens downwards, meaning it has a maximum. The maximum occurs at $k=\frac{n}{2}$. If $n$ is even, then $k=\frac{n}{2}$ is the solution. In our case, $n=100$, and thus $k=50, a_{k}=2501$ is the (only) largest value through which the diagonal passes.

If $n$ is odd, $\frac{n}{2}$ is not an integer, so the maximum must be found among the closest integers; from the previous transformations, it is clear that both $\left[\frac{n}{2}\right]$ and $\left[\frac{n}{2}\right]+1$ are valid, meaning there are two largest values in this case.

Thiry Gábor (Érd, Horváth A. Gimn., 12th grade)