We are given that $ N\equiv 5\pmod{8} $ and $ N\equiv 3\pmod{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\pmod{8} $ and $ x\equiv 3\pmod{6} $ are those of the form $ x=21+\text{lcm}(6,8)k = 21 + 24 k $, 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} $.