Problem 2374

Given an integer $n \ge 1$, we need to find the smallest possible value of
$$\sum_{k=1}^{n}\left\lfloor\frac{a_k}{k}\right\rfloor$$
over all permutations $(a_1, a_2, \ldots, a_n)$ of $\{1, 2, \ldots, n\}$.

### Step-by-step Approach

1. Understanding the Floor Function: The expression $\left\lfloor\frac{a_k}{k}\right\rfloor$ returns the greatest integer less than or equal to $\frac{a_k}{k}$. This function decreases the output by approximately $k-1$ each time $a_k$ increases by $k$.

2. Goal Analysis: Our aim is to minimize the sum of all such floor expressions for every $k$ in $[1, n]$. Attempting to place each $a_k$ such that $\frac{a_k}{k}$ is minimized will generally result in minimizing the entire sum.

3. **Strategic Choice of $a_k$:
- Incremental Strategy**:
- Start with smaller numbers for smaller divisions.
- For each $k$, we ideally want the smallest value of $a_k$ such that $\left\lfloor \frac{a_k}{k} \right\rfloor$ yields the minimum value $0$.
- If $a_k \leq k$, then $\left\lfloor \frac{a_k}{k} \right\rfloor = 0$.
- Hence, for $k$ up to a certain threshold, the elements should be within $\{1, 2, \ldots, k\}$.

4. Determining the Threshold:
- The transition from $0$ to $1$ in $\left\lfloor \frac{a_k}{k} \right\rfloor$ happens when $a_k = k+1$. To maintain a $0$ value, choose $a_k \leq k$.
- The smallest $k$ beyond which we can't maintain $a_k \leq k$ perfectly over all indices is determined by balancing allocations in available slots.

5. Logarithmic Insight:
- Arrange permutations such that the maximum $a_k$ aligns conceptually with binary powers due to the floor function behavior which mirrors log curves.
- When $k$ is a power of 2, new allocations necessitate increased $\left\lfloor \cdot \right\rfloor$ values.

6. Realizing the Sum:
- Each successive block (doubling size of indices) achieves at least an increase by $1$.
- Thus, iterating through powers of 2 economically, the entire strategy aligns with $\left\lfloor \log_2(n) \right\rfloor + 1$, due to the additive nature of each power block.

Consequently, considering blocks of decreasing increments, balancing partitions as suggested aligns the structural floor contributions with logarithmic expectations.

### Conclusion

The smallest possible value of the sum for any permutation $(a_1, a_2, \ldots, a_n)$ is therefore:

$$\boxed{\lfloor \log_2(n) \rfloor + 1}$$

This minimizes accumulated floor term contributions across all examined permutations.