Problem 1585

1. Assumption and Simplification:
- We start by noting that the sequence $a_1, a_2, \ldots, a_n$ satisfies the given conditions if and only if the sequence $(a_1 + w, a_2 + w, \ldots, a_n + w)$ also satisfies the conditions for any real number $w$. Therefore, we can assume without loss of generality that $a_1 = 0$.

2. Subset and Condition:
- Consider any $m$-element subset $I = \{i_1, i_2, \ldots, i_m\}$ of $\{1, 2, \ldots, n\}$ with $i_1 < i_2 < \cdots < i_m$. We need to ensure that for any such subset, if $\sum_{i \in I} i > k$, then $\frac{1}{m} \sum_{i \in I} a_i > \frac{1}{n} \sum_{i=1}^n a_i$.

3. Rewriting the Condition:
- Given $a_1 \leq a_2 \leq \cdots \leq a_n$, we can rewrite the sequence as $(0, 0, \ldots, 0, a_{i_1}, a_{i_1}, \ldots, a_{i_1}, a_{i_2}, a_{i_2}, \ldots, a_{i_2}, \ldots, a_{i_m}, a_{i_m}, \ldots, a_{i_m})$.

4. Inequality Transformation:
- We need to show that:
$$\frac{a_{i_1} + a_{i_2} + \cdots + a_{i_m}}{m} \geq \frac{\left( \sum_{j=1}^{m-1} a_{i_j} (i_{j+1} - i_j) \right) + a_{i_m} (n - i_m + 1)}{n}$$
Simplifying, we get:
$$\left( \sum_{j=1}^{m-1} a_{i_j} (n - m (i_{j+1} - i_j)) \right) + a_{i_m} (n - m (n - i_m + 1)) \geq 0$$

5. Introducing Differences:
- Let $t_{i_k} = a_{i_k} - a_{i_{k-1}} > 0$ for $k = 2, 3, \ldots, m$. The inequality reduces to:
$$a_{i_1} (mn - m (n - i_1 + 1)) + \sum_{j=2}^{m} t_{i_j} ((m - (j-1)) n - m (n - i_j + 1)) \geq 0$$
where $t_{i_2}, t_{i_3}, \ldots, t_{i_m} > 0$ and $a_{i_1} \geq 0$.

6. Non-existence Condition:
- There does not exist such $a_{i_1}, t_{i_1}, t_{i_2}, \ldots, t_{i_m}$ if and only if:
$$mn - m (n - i_1 + 1) < 0 \quad \text{and} \quad (m - (j-1)) n - m (n - i_j + 1) < 0 \quad \text{for all } j = 2, 3, \ldots, m$$

7. Choosing Indices:
- This means that if $i_1 = 1 = \left\lfloor \frac{n(1-1)}{m} + 1 \right\rfloor$ and $i_j = \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor$ for all $j = 2, 3, \ldots, m$, then there does not exist $a_1, a_2, \ldots, a_n$.

8. **Finding $k$**:
- Therefore, $k \geq 1 + \sum_{j=2}^{m} \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor$.
- For these values of $k$, we get that $\sum_{j=1}^{m} \left( i_j - \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor \right) > 0$.
- This implies there exists $h \in \{1, 2, \ldots, m\}$ such that $i_h > \left\lfloor \frac{n(h-1)}{m} + 1 \right\rfloor$, then choosing $t_{i_h} \rightarrow \infty$ makes the inequality hold.

9. **Smallest Value of $k$**:
- Hence, the smallest value of $k$ is:
$$\sum_{j=1}^{m} \left( \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor \right) = \sum_{j=1}^{m} \left( 1 + \frac{n(j-1)}{m} \right) - \left( \frac{0}{m} + \frac{1}{m} + \cdots + \frac{m-1}{m} \right) = \frac{(n-1)(m-1)}{2} + m$$

The final answer is $\boxed{\frac{(n-1)(m-1)}{2} + m}$