The set $S=\{1,2,\dots,2020\}$ admits the partition $S=\cup_{m=1}^{1000} S_m$, where $S_m=\{(2m-1)\cdot2^0,(2m-1)\cdot2^1,\dots,(2m-1)\cdot2^k\};k\in\mathbb{Z};(2m-1)\cdot2^k\le 2000<(2m-1)\cdot2^{k+1}$. Denote $L_m=|S_m|=\left\lfloor\log_2\left(\dfrac{2000}{2m-1}\right)\right\rfloor+1$. The minimum number of marked numbers in the set $S_m$ is $\left\lfloor\dfrac{L_m}{2}\right\rfloor$, respectively the numbers $(2m-1)\cdot2^b$, where $b$ is odd. Hence: $\min N=\sum_{m=1}^{1000}\left\lfloor\dfrac{L_m}{2}\right\rfloor$. $\max_{1\le m\le1000}L_m=L_1=\lfloor\log_22000\rfloor+1=11$. We obtain successively: $L_m=11\Longrightarrow m=1$, hence $1$ possible value of $m$; $L_m=10\Longrightarrow m=2$, hence $1$ possible value of $m$; $L_m=9\Longrightarrow m\in\{3,4\}$, hence $2$ possible values of $m$; $L_m=8\Longrightarrow m\in\{5,6,7,8\}$, hence $4$ possible values of $m$; $L_m=7\Longrightarrow 9\le m\le 16$, hence $8$ possible values of $m$; $L_m=6\Longrightarrow 17\le m\le 31$, hence $15$ possible values of $m$; $L_m=5\Longrightarrow 32\le m\le 63$, hence $32$ possible values of $m$; $L_m=4\Longrightarrow 64\le m\le 125$, hence $62$ possible values of $m$; $L_m=3\Longrightarrow 126\le 250$, hence $125$ possible values of $m$; $L_m=2\Longrightarrow 251\le 500$, hence $250$ possible values of $m$; $L_m=1\Longrightarrow 501\le m\le1000$, hence $500$ possible values of $m$. Results: $\min N=1\cdot\left\lfloor\dfrac{11}{2}\right\rfloor+1\cdot\left\lfloor\dfrac{10}{2}\right\rfloor+2\cdot\left\lfloor\dfrac{9}{2}\right\rfloor+4\cdot\left\lfloor\dfrac{8}{2}\right\rfloor+8\cdot\left\lfloor\dfrac{7}{2}\right\rfloor+15\cdot\left\lfloor\dfrac{6}{2}\right\rfloor+$ $+32\cdot\left\lfloor\dfrac{5}{2}\right\rfloor+62\cdot\left\lfloor\dfrac{4}{2}\right\rfloor+125\cdot\left\lfloor\dfrac{3}{2}\right\rfloor+250\cdot\left\lfloor\dfrac{2}{2}\right\rfloor+500\cdot\left\lfloor\dfrac{1}{2}\right\rfloor=666$.