jmc

Let a common solution be $a$. Then we know $$\begin{array}{rl}a\equiv 3& (\mathrm{mod}4)\\ a\equiv 1& (\mathrm{mod}3)\\ a\equiv 1& (\mathrm{mod}5)\end{array}$$ Since $\gcd (3,5)=1$, $(2)$ and $(3)$ together yield $a\equiv 1(\mathrm{mod}3\cdot 5)$ which is the same as $a\equiv 1(\mathrm{mod}15)$. Then there exists an integer $n$ such that $a=1+15n$. Substituting this into $(1)$ gives $$1+15n\equiv 3(\mathrm{mod}4)⟹n\equiv 2(\mathrm{mod}4)$$ So $n$ has a lower bound of $2$. Then $n\ge 2⟹a=1+15n\ge 31$. $31$ satisfies the original congruences so subtracting it from both sides of each gives $$\begin{array}{rl}a-31\equiv -28\equiv 0& (\mathrm{mod}4)\\ a-31\equiv -30\equiv 0& (\mathrm{mod}3)\\ a-31\equiv -30\equiv 0& (\mathrm{mod}5)\end{array}$$ Since $\gcd (3,4)=\gcd (4,5)=\gcd (3,5)=1$, we have $a-31\equiv 0(\mathrm{mod}3\cdot 4\cdot 5)$, that is, $a\equiv 31(\mathrm{mod}60)$.

Note that any solution of the above congruence also satisfies the original ones. Then the two solutions are $31$ and $60+31=91$. Thus, $31+91=\boxed{122}$.