Since $a$ is a positive integer and $a + \frac{b}{c}$ is a positive integer, then $\frac{b}{c}$ is a positive integer. In other words, $b$ is a multiple of $c$. Similarly, since $\frac{a}{c} + b$ is a positive integer and $b$ is a positive integer, then $a$ is a multiple of $c$. Thus, we can write $a = Ac$ and $b = Bc$ for some positive integers $A$ and $B$. Therefore, $a + \frac{b}{c} = 101$ becomes $Ac + B = 101$ and $\frac{a}{c} + b = 68$ becomes $A + Bc = 68$. Adding these new equations gives $Ac + B + A + Bc = 101 + 68$ or $A(c + 1) + B(c + 1) = 169$ and so $(A + B)(c + 1) = 169$. Since $(A + B)(c + 1) = 169$, then $c + 1$ is a divisor of 169. Since $169 = 13^{2}$, then the positive divisors of 169 are $1, 13, 169$. Since $A, B, c$ are positive integers, then $A + B \geq 2$ and $c + 1 \geq 2$. Since neither $A + B$ nor $c + 1$ can equal 1, then $A + B = c + 1 = 13$. Finally, $\frac{a + b}{c} = \frac{Ac + Bc}{c} = A + B = 13$ and so $k = 13$.