The set $S=\{(x,y) \in \mathbb{R}^{n} \times \mathbb{R}^n = \mathbb{R}^{2n} ; x \neq y\}$ is connected if $n \geq 2$. When n = 1 it is easy to see that is not connected, it just take the split open $ S=U_1 \cup U_2$ such that $U_1 = \{(x,y) \in \mathbb{R}^2 ; x > y\}$ is $U_2 = \{(x,y) \in \mathbb{R}^2 ; x < y\}$ not trivial.
 A: Up to a rotation of $\mathbb R ^{2n}$ you ask for which $n$ the space $\mathbb R^{2n} - \mathbb R^{n}$ is path connected, where $\mathbb R^{n}$ sits in the first $n$ coordinates. But contracting the first $n$ coordinates is a homotopy equivalence ending in $\mathbb R^{n} - 0$.
A: Hint: Let $n=2$. Imagine you have $2$ pairs of points in a plane (the coordinates of $2$ points in $\mathbb R^{2\cdot2}$). Can you transform one pair into the other without collapsing it to a point anywhere along the way?

Answer: Consider two pairs of point $(x_1,y_1), (x_2,y_2)$.
If we find a continuous transformation $\varphi(t):[0,1]\to\mathbb R^n\times\mathbb R^n$, $\varphi(t)=(\varphi_1(t),\varphi_2(t))$ such that $\varphi(0)=(x_1,y_1)$, $\varphi(1)=(x_2,y_2)$ and $\varphi_1(t)\ne\varphi_2(t)$ for all $t\in[0,1]$, then $\varphi$ determines an arc in $S=R^{2n}\setminus\{(x,x):x\in\mathbb R^n\}$ connecting the points $(x_1,y_1)$ and $(x_2,y_2)$.
It's easy to see there are many ways to do it. E.g. in $n=2$ we can always connect the points like this and obviously $\varphi_1(t)\ne\varphi_2(t)$, because the arcs don't even intersect.

In a formal proof, you would use polygonal lines and you should do that yourself.
A: Lemma 1. If $x,y,u,v\in\mathbb R^n$ are such that  $x-y,u-v$ are linearly independent, then the straight line joining $(x,y)$ and $(u,v)$ is a subset of $S$.
Proof: The general point on the line is of the form $t(x,y)+(1-t)(u,v)$. If such a point is $\notin S$, then $tx+(1-t)u=ty+(1-t)v$, i.e. $t(x-y)+(1-t)(u-v)=0$. By linear independence, $t=1-t=0$, contradiction. $_\square$
Lemma 2. If $V$ is a vector space,  $u,v,w\in V$ and $u,v$ are linearly independent, and $u,w$ are linearly dependent, and $v,w$ are linearly dependent, then $w=0$. 
Proof: Assume $au+bw=cv+dw=0$ with $(a,b)\ne 0$ and $(c,d)\ne 0$. If $a=0$ or $c=0$ we are done. From $0=d(au+bw)-b(cv+dw)=adu-bcv$, we conclude $ad=bc=0$ and then $d=b=0$. But then $au=cv$. $_\square$
Proposition. If $u,v\in\mathbb R^n$ are linearly independent and $(x,y)\in S$, then one of the straight lines connecting $(x,y)$ with $(0,u)$ or with $(0,v)$ is s subset of $S$.
Proof: Let $w=x-y$ in lemma 2. As $(x,y)\in S$, w have $w\ne 0$, hence $u,w$ or $v,w$ are linearly independent. Then the claim follows form lemma 1. $_\square$
Corollary. If $n\ge 2$ then $S$ is path connected.
Proof: We can find two linearly independent vectors $u,v$ in $\mathbb R^n$.
By lemma 1, the straight line from $(0,u)$ to $(0,v)$ is in $S$ and any point $(x,y)\in S$ is path connected to one of these two points by the proposition. $_\square$
