The answer is no. E.g., assume that $n=2$, $\xi_i=X_i$, $P(X_i=0)=P(X_i=1)=1/2$, and $X_1,X_2$$X_1,X_2,\nu$ are independentany random variables (r. Letv's) each with values in the set $\nu$ equal$\{1,2\}$ and with $1$ if$P(\nu=1)=p\in(0,1)$. With these conditions in place, the dependence between $X_1\le X_2$ and let$X_1,X_2,\nu$ may be arbitrary. For instance, we may suppose that $\nu$ equalis independent of $2$ if$X_1,X_2$, or we may suppose that $X_1>X_2$$\nu=X_1$, or . Let.. .
Take any $c\notin\{1,2\}$ and let $Y_1,Y_2$ be obtained from $X_1,X_2$ by replacing $X_\nu$ by $c\notin\{0,1\}$. Then$c$, so that
$$P(Y_1=c,Y_2=c)=0\ne\frac34\times\frac14=P(Y_1=c)P(Y_2=c).
$$\begin{equation}
(Y_1,Y_2)=\left\{
\begin{aligned}
(c,X_2) &\text{ if }\nu=1,\\
(X_1,c) &\text{ if }\nu=2.
\end{aligned}
\right.
\end{equation}
Then
$$P(Y_1=c,Y_2=c)=0\ne p(1-p)=P(\nu=1)P(\nu=2)=P(Y_1=c)P(Y_2=c).
$$
So, $Y_1,Y_2$ are not independent.