One, (20 points) Real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Find the maximum and minimum values of $x+y+z+w$.
Let $z=w+a, y=w+a+b, x=w+$ $a+b+c$. Then $a, b, c \geqslant 0$, and
$$
x+y+z+w=4 w+3 a+2 b+c .
$$

Therefore, $100=5(w+a+b+c)+4(w+a+$
$$
\begin{array}{l}
\quad b)+3(w+a)+6 w \\
=18 w+12 a+9 b+5 c \\
=4(4 w+3 a+2 b+c)+(2 w+b+c) \\
\geqslant 4(x+y+z+w) .
\end{array}
$$

Thus, $x+y+z+w \leqslant 25$.
When $x=y=z=\frac{25}{3}, w=0$, the equality holds. Therefore, the maximum value of $x+y+z+w$ is 25.
$$
\begin{array}{l}
\text { Also, } 100=18 w+12 a+9 b+5 c \\
=5(4 w+3 a+2 b+c)-(2 w+3 a+b) \\
\leqslant 5(x+y+z+w),
\end{array}
$$

then $x+y+z+w \geqslant 20$.
When $x=20, y=z=w=0$, the equality holds. Therefore, the minimum value of $x+y+z+w$ is 20.