Ones and Zeros

$1+11+101+1001+ \cdots +1 \underbrace{0 \cdots 0}_{50}1 = 1+(10+1)+(100+1)+(1000+1)+ \cdots + (1 \underbrace{0 \cdots 0}_{51} +1) = \underbrace{11 \cdots 11}_{51}0 + 52 = \underbrace{11 \cdots 11}_{50}62$ .

$50 \times 1 +6+2 = \boxed{58}$ .

$\begin{aligned} N & = 1 + 11 + 101+ 1001+ 10001+... + 1\overbrace{000...000}^{\text{50 zeros}}1 \\ & = \sum_{n=0}^{51} \left(10^n + 1\right) - 1 \\ & = \frac {10^{52}-1}{10-1} + 52 - 1 \\ & = \frac {\overbrace{999...999}^{\text{52 nines}}} 9 + 51 \\ & = \overbrace{111...111}^{\text{52 ones}} + 51 \\ & = \overbrace{111...111}^{\text{50 ones}} 62 \end{aligned}$

Therefore, the sum of the digits of $N$ is $50 \times 1 + 6 + 2 = \boxed{58}$ .