The independence will be then in general lost. E.g., let $X_1,\dots,X_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X_1,\dots,X_n)=X_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y_1,\dots,Y_{n-1})$ is the leftover sequence, after the removal of $X_\nu$. Then, conditionally on $M$, the $Y_i$'s are iid uniformly distributed on $[0,M]$.

So, for $n\ge2$ and $i\in[2]$ we have $E(Y_i|M)=M/2$ and $E(Y_1Y_2|M)=E(Y_1|M)E(Y_2|M)=(M/2)^2$. So, $$EY_1Y_2=E(M/2)^2>(EM/2)^2=EY_1\,EY_2,$$ with the inequality taking place because $Var\,M>0$. Thus, $Y_1$ and $Y_2$ are not independent.

Intuitively, it can be expected that the $Y_i$'s are positively dependent. Indeed, if $M$ is small, then all $Y_i$'s will be small. So, if $Y_1$ turns out to be small, a reason for that may be that $M$ is small, and then $Y_2$ will be small. Thus, the smallness of $Y_1$ seems to make $Y_2$ tend to be small.

The independence will be then in general lost. E.g., let $X_1,\dots,X_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X_1,\dots,X_n)=X_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y_1,\dots,Y_{n-1})$ be the leftover sequence, after the removal of $X_\nu$. Then, conditionally on $M$, the $Y_i$'s are iid uniformly distributed on $[0,M]$.

So, for $n\ge2$ and $i\in[2]$ we have $E(Y_i|M)=M/2$ and $E(Y_1Y_2|M)=E(Y_1|M)E(Y_2|M)=(M/2)^2$. So, $$EY_1Y_2=E(M/2)^2>(EM/2)^2=EY_1\,EY_2,$$ with the inequality taking place because $Var\,M>0$. Thus, $Y_1$ and $Y_2$ are not independent.

Intuitively, it can be expected that the $Y_i$'s are positively dependent. Indeed, if $M$ is small, then all $Y_i$'s will be small. So, if $Y_1$ turns out to be small, a reason for that may be that $M$ is small, and then $Y_2$ will be small. Thus, the smallness of $Y_1$ seems to make $Y_2$ tend to be small.

The independence will be then in general lost. E.g., let $X_1,\dots,X_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X_1,\dots,X_n)=X_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y_1,\dots,Y_{n-1})$ is the leftover sequence, after the removal of $X_\nu$. Then, conditionally on $M$, the $Y_i$'s are iid uniformly distributed on $[0,M]$.

So, for $n\ge2$ and $i\in[2]$ we have $E(Y_i|M)=M/2$ and $E(Y_1Y_2|M)=E(Y_1|M)E(Y_2|M)=(M/2)^2$. So, $$EY_1Y_2=E(M/2)^2>(EM/2)^2=EY_1\,EY_2,$$ with the inequality taking place because $Var\,M>0$. Thus, $Y_1$ and $Y_2$ are not independent.

The independence will be then in general lost. E.g., let $X_1,\dots,X_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X_1,\dots,X_n)=X_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y_1,\dots,Y_{n-1})$ is the leftover sequence, after the removal of $X_\nu$. Then, conditionally on $M$, the $Y_i$'s are iid uniformly distributed on $[0,M]$.

So, for $n\ge2$ and $i\in[2]$ we have $E(Y_i|M)=M/2$ and $E(Y_1Y_2|M)=E(Y_1|M)E(Y_2|M)=(M/2)^2$. So, $$EY_1Y_2=E(M/2)^2>(EM/2)^2=EY_1\,EY_2,$$ with the inequality taking place because $Var\,M>0$. Thus, $Y_1$ and $Y_2$ are not independent.

Intuitively, it can be expected that the $Y_i$'s are positively dependent. Indeed, if $M$ is small, then all $Y_i$'s will be small. So, if $Y_1$ turns out to be small, a reason for that may be that $M$ is small, and then $Y_2$ will be small. Thus, the smallness of $Y_1$ seems to make $Y_2$ tend to be small.