The product of a million whole numbers is equal to million. What can be the greatest possible value of the sum of these numbers?
1. **Claim**: The maximal sum \(a + b\) with a fixed product \(ab\) occurs when one of the two numbers is \(1\).

2. **Proof**: Let the fixed product \(ab\) be \(x\). We desire to show that \(x + 1 > p + q\), where \(p, q > 1\) and \(pq = x\). This is obvious because since \(p, q > 1\), \((p-1)(q-1) > 0\), which rearranges to \(pq + 1 > p + q\), or \(x + 1 > p + q\).

3. **Application to the problem**: Now consider an arrangement of the million whole numbers with the maximal possible sum. If any two terms \(a_i\) and \(a_j\) aren't ones, we can change them to \(1\) and \(a_i a_j\), to keep the overall product the same. From our claim, this increases the total sum, contradicting our assumption that the arrangement had the maximal possible sum. Therefore, one cannot find two terms \(a_i\) and \(a_j\) that aren't ones, meaning at most one number isn't a one.

4. **Conclusion**: This means that our optimizing arrangement has to be \(1000000, 1, 1, \cdots\), meaning the greatest possible sum is:
   \[
   1000000 + 1 \cdot (1000000 - 1) = 1000000 + 999999 = 1999999
   \]

The final answer is \(\boxed{1999999}\).