$解:(1)原式可化為|4-2m|+4-2m+(n-2)2$
$+ \sqrt{(m-2)n2} =0.\ $
$∵m-2≥0，∴m≥2，∴4-2m≤0.$
$∴原式可化為(n-2)2+ \sqrt{(m-2)n2} =0.$
$∵(n-2)2≥0， \sqrt{(m-2)n2} ≥0.$
$∴\begin{cases}{ n-2=0, }\ \\ {\ \sqrt {(m-2)n2}=0,} \end{cases} 即\begin{cases}{\ n=2,}\ \\ {m=2,\ } \end{cases}\ $
$∴m+n=2+2=4.$
$(2)根據(jù)題意，得\begin{cases}{ x-2023+y≥0, }\ \\ {\ 2023-x-y≥0,} \end{cases}\ $
$則x+y=2023，$
$即 \sqrt{3x+2y-1-m}+\sqrt{2x+3y-m}=0.$
$∴\begin{cases}{ x+y=2023, }\ \\ { 3x+2y-1-m=0, }\\{2x+3y-m=0,} \end{cases} 解得\begin{cases}{ x=1012, }\ \\ { y=1011, }\\{m=5057,} \end{cases}\ $
$故m=5057.$
