We know that \log_2(x) increases on (0, +\infty). Also, \log_2(2^n)=n, and there are 2 numbers from 2 to 3, 4 numbers from 4 to 7, ..., 128 numbers from 128 to 255, hence \sum_{n=1}^256\lfloor\log_2(n)\rfloor= 0+2 \cdot 1+ 4 \cdot 2 + 8 \cdot 3 + \cdots + 128 \cdot 7 + 8 = 1546. Sorry for inconvenience, but since I am new here so I can’t send images which stops me from using LaTeX.

$LaTeXed!$ We know that : $\log_2 x$ increases on $(0, +\infty).$ Also, $\log_2 2^n=n,$ and there are $2$ numbers from $2$ to $3,4$ numbers from $4$ to $7,\cdots,128$ numbers from $128$ to $255,$ hence $\sum_{n=1}^{256} \lfloor\log_2 n\rfloor= 0+2 \cdot 1+ 4 \cdot 2 + 8 \cdot 3 + \cdots + 128 \cdot 7 + 8 = 1546.$