1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$

can take  different values.
$=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)$. The first term on the right side, $\left(x^{2}+y^{2}+z^{2}+t^{2}\right)$, is also fully symmetric, so $n$ takes different values only depending on $x z+y t$, which has exactly three different values: $a b+c d, a c+b d, a d+b c$. In fact,
$$
\begin{array}{l}
(a b+c d)-(a c+b d)=a(b-c)+d(c-b) \\
=(b-c)(a-d)>0 .
\end{array}
$$

Thus, $a b+c d>a c+b d$.
Similarly, $a c+b d>a d+b c$.
Therefore, $n(x, y, z, t)$ can take three different values.