How is this possible?

Let $g(n)$ be the largest factor of $n$ less than or equal to $\sqrt{n}$ . $f(n) = n/g(n).$

$\sum_{n=1}^{m^2} f(n) = \sum_{n=1}^{m^2} n / g(n) = \sum_{k=1}^m \frac{1}{k} \sum_{n : k^2 \leq n \leq m^2 \\ \text{ AND } g(n)=k} n = \sum_{k=1}^m \sum_{p : p \text{ prime } \\ \text{AND } k < p \leq m^2/k} p + \sum_{k=1}^m \frac{1}{k} \sum_{n : k^2 \leq n \leq m^2 \\ \text{ AND } g(n)=k \\ \text{ AND } n \neq k p \\ \text{ for any prime } p > k} n.$

It is easy to prove that if $g(n) = k$ and $k^2 \leq n \leq m^2 \text{ and } n \neq k p \text{ for any prime } p > k$ , then $n$ is a product of primes each less than or equal to $k$ . You can use this to show that

$\sum_{k=1}^m \frac{1}{k} \sum_{n : k^2 \leq n \leq m^2 \\ \text{ AND } g(n)=k \\ \text{ AND } n \neq k p \\ \text{ for any prime } p > k} n = o\left(\frac{m^4}{\log{m^2}}\right).$

Using that $\sum_{p \leq x} p \sim \frac{x^2}{2 \log{x}}$ ,

$\sum_{k=1}^m \sum_{p : p \text{ prime AND } \\ k < p \leq m^2/k} p \sim \frac{\left(m^2\right)^2}{2 \log \left(m^2\right)} + \sum_{k=2}^m \left(\frac{\left(\frac{m^2}{k}\right )^2}{2 \log \left(\frac{m^2}{k}\right)}-\frac{k^2}{2 \log (k)}\right).$

So

$\frac{\log{m^2}}{m^4} \sum_{n=1}^{m^2} f(n) = \frac{\log{m^2}}{m^4} \sum_{k=1}^m \sum_{p : p \text{ prime AND } \\ k < p \leq m^2/k} p + o(1) = \frac{1}{2} \left(\sum _{k=2}^m \left(\frac{\left(\frac{m^2}{k}\right)^2 \log \left(m^2\right)}{m^4 \log \left(\frac{m^2}{k}\right)}-\frac{k^2 \log \left(m^2\right)}{m^4 \log (k)}\right)+1\right) + o(1) =\frac{1}{2} \left(\sum _{k=2}^m \left(\frac{\log (k)}{\left(2 k^2\right) \log (m)}+\frac{1}{k^2}+o\left(\frac{1}{\log (m)}\right)\right)+1\right) + o(1)=\frac{1}{2} \left(\sum _{k=2}^m \frac{1}{k^2}+1\right)=\frac{\pi ^2}{12} + o(1)$

The penultimate equality requires rigorous justification which I omit.

So the answer is $2+12=14$ .

This method is cumbersome. Does anybody have a better answer? It feels like something simpler is hiding under all this ...