smc

We start with $2^{1+\lfloor {\log }_2(N-1)\rfloor }-N=19$. After rearranging, we get $\lfloor {\log }_2(N-1)\rfloor ={\log }_2\left(\frac{N+19}{2}\right)$. Since $\lfloor {\log }_2(N-1)\rfloor$ is a positive integer, $\frac{N+19}{2}$ must be in the form of $2^m$ for some positive integer $m$. From this fact, we get $N=2^{m+1}-19$. If we now check integer values of N that satisfy this condition, starting from $N=19$, we quickly see that the first values that work for $N$ are $45$ and $109$, that is, $2^6-19$ and $2^7-19$, giving values of $5$ and $6$ for $m$, respectively. Adding up these two values for $N$, we get $45+109=154\to \boxed{154}$