Maybe I am wrong...

Let $G=\gcd(a,b)$ and $L = \mbox{lcm}(a,b)$ . Then we have:

$\begin{aligned} \gcd(a,b) + \mbox{lcm}(a,b) + \color{#3D99F6} ab & = 2014 & \small \color{#3D99F6} \text{Note that }\gcd(a,b) \times \mbox{lcm}(a,b) = ab \\ G + L + GL & = 2014 \\ GL + G + L +1 & = 2015 \\ (G+1)(L+1) & = 2015 & \small \color{#3D99F6} \text{and that }2015 = 5 \times 13 \times 31 \end{aligned}$

Since $G \le L$ , the possible values of $G$ and $L$ are $\begin{cases} 1 \times 2015 = 2015 & \implies G= 0 & L = 2014 \\ 5 \times 403 = 2015 & \implies G= 4 & L = 402 \\ 13 \times 155 = 2015 & \implies G= 12 & L = 154 \\ 31 \times 65 = 2015 & \implies G= 30 & L = 64 \end{cases}$

We note that $(G, L) = (0, 2014)$ is unacceptable. Since $G^2 \mid ab$ , we have $G^2 \mid GL \implies G \mid L$ . We note that none of the remaining three $L$ 's us divisible by the respective $G$ 's. Come on Frank! There is no $a$ and $b$ that satisfy this!