Problem 1526

1. Assume the contrary: Suppose that $a_{m,n} \leq mn$ for all $m, n \in \mathbb{N}$.

2. Count the number of distinct values: For a fixed $K \in \mathbb{N}$, consider the numbers $a_{m,n}$ with $mn \leq K$. Since $a_{m,n} \leq mn$, these numbers can take at most $K$ distinct values.

3. Use the given condition: Each positive integer appears exactly 8 times in the array. Therefore, there can be at most $8K$ pairs $(m,n)$ with $mn \leq K$.

4. **Count the pairs $(m,n)$ with $mn \leq K$**: For each $n \leq K$, there are exactly $\left\lfloor \frac{K}{n} \right\rfloor$ possible values of $m$ such that $mn \leq K$. Therefore, the total number of such pairs is:
$$\sum_{n=1}^K \left\lfloor \frac{K}{n} \right\rfloor$$

5. Estimate the sum: We can estimate the sum using the inequality $\left\lfloor \frac{K}{n} \right\rfloor \geq \frac{K}{n} - 1$:
$$\sum_{n=1}^K \left\lfloor \frac{K}{n} \right\rfloor \geq \sum_{n=1}^K \left( \frac{K}{n} - 1 \right) = K \sum_{n=1}^K \frac{1}{n} - K$$

6. Simplify the sum: The sum can be further simplified as:
$$K \sum_{n=1}^K \frac{1}{n} - K = K \left( \sum_{n=1}^K \frac{1}{n} - 1 \right) = K \sum_{n=2}^K \frac{1}{n}$$

7. Use the harmonic series: Since the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, for large enough $K$, $\sum_{n=2}^K \frac{1}{n}$ can be made arbitrarily large.

8. Contradiction: This implies that for large enough $K$, the number of pairs $(m,n)$ with $mn \leq K$ exceeds $8K$, which contradicts the assumption that there are at most $8K$ such pairs.

9. Conclusion: Therefore, the assumption that $a_{m,n} \leq mn$ for all $m, n \in \mathbb{N}$ must be false. Hence, there exists some pair of positive integers $(m, n)$ such that $a_{m,n} > mn$.

$\blacksquare$