Does there exist a nonnegative integer $a$ for which the equation 
\[\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a\]
has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers?

[i]The expression $\lfloor x\rfloor$ denotes the integer part (or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor = 1, \lfloor\pi\rfloor =\lfloor 22/7 \rfloor = 3, \lfloor 42\rfloor = 42,$ and $\lfloor 0 \rfloor = 0$.[/i]

To determine if there exists a nonnegative integer \( a \) such that the equation 
\[
\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a
\]
has more than one million different solutions \((m, n)\) where \( m \) and \( n \) are positive integers, we will proceed as follows:

1. **Define \( S_m \):**
   Let \( S_m = \sum_{i=1}^m \left\lfloor \frac{m}{i} \right\rfloor \). We need to analyze the behavior of \( S_m \).

2. **Double Counting Argument:**
   We will use a double counting argument to show that:
   \[
   S_m = \sum_{i=1}^m d(i)
   \]
   where \( d(i) \) is the number of divisors of \( i \). This is because each divisor \( i \) of \( m \) contributes to the sum \( \left\lfloor \frac{m}{i} \right\rfloor \).

3. **Bounding \( d(i) \):**
   We know that \( d(i) \leq 2\sqrt{i} \) for all \( i \in \mathbb{N} \). This is because the number of divisors of \( i \) is at most twice the square root of \( i \).

4. **Bounding \( S_m \):**
   Using the bound on \( d(i) \), we get:
   \[
   S_m = \sum_{i=1}^m d(i) \leq \sum_{i=1}^m 2\sqrt{i}
   \]
   Using the integral approximation, we can bound the sum:
   \[
   \sum_{i=1}^m \sqrt{i} \approx \int_1^m \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_1^m = \frac{2}{3} (m^{3/2} - 1)
   \]
   Therefore,
   \[
   S_m \leq 2 \cdot \frac{2}{3} m^{3/2} = \frac{4}{3} m^{3/2}
   \]

5. **Choosing \( n \):**
   For each fixed \( m \), choose \( n = \lfloor \sqrt{S_m} \rfloor \). Then,
   \[
   n^2 \leq S_m < (n+1)^2
   \]
   This implies:
   \[
   n^2 \leq \frac{4}{3} m^{3/2} < (n+1)^2
   \]

6. **Bounding \( a \):**
   Since \( S_m = n^2 + a \), we have:
   \[
   a = S_m - n^2
   \]
   Given the bounds on \( S_m \) and \( n \), we get:
   \[
   0 \leq a < 2n + 1
   \]
   Since \( n \approx \sqrt{S_m} \approx \sqrt{\frac{4}{3} m^{3/2}} = \frac{2}{\sqrt{3}} m^{3/4} \), we have:
   \[
   a < 2 \cdot \frac{2}{\sqrt{3}} m^{3/4} + 1 = \frac{4}{\sqrt{3}} m^{3/4} + 1
   \]

7. **Counting Solutions:**
   For each \( m \), there are at most \( \frac{4}{\sqrt{3}} m^{3/4} + 1 \) possible values of \( a \). If there were no such \( a \) that allows more than one million solutions, then:
   \[
   \text{Number of solutions} \leq 10^6 \cdot \left( \frac{4}{\sqrt{3}} m^{3/4} + 1 \right)
   \]
   However, each \( m \) gives a unique solution for \( n \) and \( a \), implying:
   \[
   \text{Number of solutions} \geq O(m)
   \]
   This leads to a contradiction because \( O(m) \) grows faster than \( O(m^{3/4}) \).

Conclusion:
Thus, there must exist a nonnegative integer \( a \) such that the equation has more than one million different solutions \((m, n)\).

The final answer is \(\boxed{\textit{yes}}\)