Why?

Let $m=q n+r, 0 \leq r < n$ . Then we want to find all $k > 1$ such that if $\frac{k m + n}{m + k n} = \frac{n (k q+1)+k r}{n (k+q)+r} = c$ is an integer, then $r = 0$ .

Note that $k-c=\frac{(k-1) (k+1) n}{n (k+q)+r}.$

Let $g = gcd(r,n), r_1 = r/g, n_1 = n/g$ . Then $c$ is an integer iff

$(k-c)/n_1 = \frac{(k-1)(k+1)}{(k+q) n_1 + r_1} = a$

is an integer (since $gcd(n_1, (k+q) n_1 + r_1) = gcd(r_1, n_1) = 1$ ). Solving for $q$ ,

$q=\frac{-a k n_1-a r_1+k^2-1}{a n_1}.$

Thus $a | (k^2-1)$ , and $r_1 = (-1+k^2)/a \mod n_1$ . We need $gcd(r_1, n_1) = 1$ and $q>0$ . For $q>0$ , it suffices to have $n_1 < (k-1)/a$ . So to show that $k$ is NOT a number we are looking for it suffices to take a (positive) factor $a$ of $k^2-1$ and choose a positive integer $2 < n_1 < (k-1)/a$ such that $gcd((k^2-1)/a,n_1) =1$ . We can just take $a=1$ . It is easy to prove that for each $k > 6$ we can find such an $n_1$ (we can always just take a prime $n_1$ ). Furthermore, we can find examples that show that $k=4,6$ do not satisfy the property stated in the question. Finally, it is easy to prove that $k=2,3,5$ do satisfy this property (just use the logic used previously). So the answer is $2+3+5=10$ .