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

Answer: Yes.
Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$

$$
L\left(q^{3}\right)=\sum_{k=1}^{q^{3}}\left[\frac{q^{3}}{k}\right] \leq \sum_{k=1}^{q^{3}} \frac{q^{3}}{k} \leq q^{3} \cdot \sum_{k=1}^{q^{3}} \frac{1}{k} \leq q^{3} \cdot q=q^{4}
$$

Indeed, the first inequality results from $[x] \leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\sum_{k=1}^{q^{3}} \frac{1}{k}$ into several groups: For $j \geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\frac{1}{2^{j}}, \ldots, \frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1. Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.

Call an integer $m$ special, if it satisfies $1 \leq L(m) \leq q^{4}$. Denote by $g(m) \geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \leq L(m)$. Since there are more than $10^{6}$ special integers, more than $10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1). Hence, the answer to the problem is YES.