$\because$点$A$的坐标为$\left(4,2\right)$，点$O$的坐标为$\left(0,0\right)$，
$\therefore$线段$OA$的中点坐标为（$\frac{4+0}{2}$，$\frac{2+0}{2})$，即$\left(2,1\right)$，
$\because$点$C$的坐标为$\left(2,1\right)$，
$\therefore$点$C$为线段$OA$的中点.
过点$A$作$AD\bot y$轴于点$D$，如图所示.
$\because$点$A$的坐标为$\left(4,2\right)$，点$B$的坐标为$\left(0,5\right)$，
$\therefore AD=4$，$OD=2$，$OB=5$，
$\therefore BD=OB-OD=5-2=3$，
$\therefore AB=\sqrt{A{D}^{2}-B{D}^{2}}=\sqrt{{4}^{2}+{3}^{2}}=5=OB$，
$\therefore BC\bot OA$，
$\therefore$直线$y=kx+b$经过点$B\left(0,5\right)$，$C\left(2,1\right)$.
将$B\left(0,5\right)$，$C\left(2,1\right)$代入$y=kx+b$得：$\left\{\begin{array}{l}{5=b}\\{1=2k+b}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{k=-2}\\{b=5}\end{array}\right.$，
$\therefore k$的值为$-2$.
故答案为：$-2$.

$\because$点$A$的坐标为$\left(4,2\right)$，点$O$的坐标为$\left(0,0\right)$，
$\therefore$线段$OA$的中点坐标为（$\frac{4+0}{2}$，$\frac{2+0}{2})$，即$\left(2,1\right)$，
$\because$点$C$的坐标为$\left(2,1\right)$，
$\therefore$点$C$为线段$OA$的中点.
过点$A$作$AD\bot y$轴于点$D$，如图所示.
$\because$点$A$的坐标为$\left(4,2\right)$，点$B$的坐标为$\left(0,5\right)$，
$\therefore AD=4$，$OD=2$，$OB=5$，
$\therefore BD=OB-OD=5-2=3$，
$\therefore AB=\sqrt{A{D}^{2}-B{D}^{2}}=\sqrt{{4}^{2}+{3}^{2}}=5=OB$，
$\therefore BC\bot OA$，
$\therefore$直线$y=kx+b$经过点$B\left(0,5\right)$，$C\left(2,1\right)$.
将$B\left(0,5\right)$，$C\left(2,1\right)$代入$y=kx+b$得：$\left\{\begin{array}{l}{5=b}\\{1=2k+b}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{k=-2}\\{b=5}\end{array}\right.$，
$\therefore k$的值为$-2$.
故答案为：$-2$.