2018/19 AHSMC Part 1 Problem 16

Notice that for large enough $k$, $n_k = 0$ and thus only a finite number of terms in the sum $n_1 + n_2 + n_3 + \cdots$ are not equal to $0$. It can be also mentioned that $\lfloor x \rfloor$ counts the number of positive integers $m$ such that $m \le x$. Therefore, the requested sum counts the number of positive integer solutions of $$n \le \frac{2^k + 20182019}{2^{k+1}},\quad k \ge 1 \iff 2^k(2n-1) \le 20182019,\quad k \ge 1.$$ Since every positive even integer can be written uniquely as a product of $2^k$, for some integer $k \ge 1$, and an odd positive integer, the sum is equal to the number of positive even integers which are less than or equal to $20182019$, which is $10091009$. The answer is (B).