$(m,n)$

If $m=n$ , the only possibility is that they are both $1$ , and this is indeed a solution pair. Now assume that $m$ and $n$ are distinct.

For two fractions to sum to an integer, their denominators (when expressed in lowest terms) must be equal. Clearly the first fraction, $\frac{n}{m}$ , is already in lowest terms, since $m$ and $n$ are coprime. Since $m$ and $n$ are distinct, the fraction $\frac{14m}{n}$ must have shared factors in its numerator and denominator.

Let $h=\gcd(14m,n)$ . Since $m$ and $n$ are coprime, we have $h=\gcd(14,n)$ . Write $n=hn'$ and $14=hx$ , where $\gcd(n',x)=1$ .

The expression is now $\frac{n}{m}+\frac{mx}{n'}$ , where both fractions are in lowest terms. For this to be an integer, we need $m=n'$ ; but as $n'|n$ , we must also have $\gcd(m,n')=1$ , and so $m=n'=1$ .

So the expression becomes $n+\frac{14}{n}$ . For this to be an integer, $n$ has to divide $14$ . Since $n \neq m$ , we have to discount $n=1$ ; this leaves the solution pairs $(1,2),(1,7),(1,14)$ . Together with the pair $(1,1)$ these give the answer $\boxed4$ .