jmc

We are given that $N\equiv 5(\mathrm{mod}8)$ and $N\equiv 3(\mathrm{mod}6)$. We begin checking numbers which are 5 more than a multiple of 8, and we find that 5 and 13 are not 3 more than a multiple of 6, but 21 is 3 more than a multiple of 6. Thus 21 is one possible value of $N$. By the Chinese Remainder Theorem, the integers $x$ satisfying $x\equiv 5(\mathrm{mod}8)$ and $x\equiv 3(\mathrm{mod}6)$ are those of the form $x=21+\text{lcm}(6,8)k=21+24k$, where $k$ is an integer. Thus the 2 solutions less than $50$ are 21 and $21+24(1)=45$, and their sum is $21+45=\boxed{66}$.