What this N could be?

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

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

There are $52$ terms in the sum: the number $1$ the number $11$ , and the $50$ numbers starting with a $1$ , ending with a $1$ and with $1$ to $50$ zeroes in between.

The longest of these terms thus has $52$ digits, $50$ zeros and $2$ ones.

When the units digits of all $52$ terms are added up, their sum is $52,$ so the units digit of $N$ is $2$ and a $5$ carried to the tens digit.

In the tens digit, there is only $1$ non-zero digit: the $1$ in the number $11.$

Therefore, using the carry, the ten' digit of $N$ is $1+5= 6$ .

In each of positions $3$ to $52$ from the right-hand end, there is only one non-zero digit, which is a $1$ .

Therefore, the digit in each of these positions in N is also a $1$ .

There is no carrying to worry about.

Thus, $N = 11 · · · 1162$ , where $N$ has $52 - 2 = 50$ digits equal to $1$

The sum of the digits of $N$ is $50(1)+6+2=58$ .