If you are serious about this, search the web for the "edge-isoperimetric problem
for the grid graph". If you just want a (relatively) short solution to your specific problem, consider the following.

Let $k:=\sqrt n$, and assume for simplicity that $k$ is an integer and
$|V_1|=n/4=k^2/4$. Let $x_1,\ldots,x_k$ and $y_1,\ldots,y_k$ be the number of
points from $V_1$ on the "horizontal" and "vertical" segments of your grid,
respectively. Thus, $0\le x_i,y_j\le k$, and
$$ x_1+\dotsb+x_k=y_1+\dotsb+y_k=k^2/4. \tag{1} $$
Write $X:=\max x_i,\ \xi:=\min x_i,\ Y:=\max y_i$, and $\eta:=\min y_i$.
Also, denote by $a$ the number of those indices $i\in[1,k]$ with $0<x_i<k$,
and, similarly, denote by $b$ the number of those $i\in[1,k]$ with $0<y_i<k$.
Finally, let $\partial(V_1)$ be the set of all those edges joining a vertex
from $V_1$ with a vertex from $V_2$. We want to show that $|\partial(V_1)|\ge
k$.

For every $i\in[1,k]$ with $0<x_i<k$, the $i$th horizontal segment
contributes at least one edge to $\partial(V_1)$; hence, the number of
horizontal edges in $\partial(V_1)$ is at least $a$. Also, for each
$i\in[1,k-1]$ the number of horizontal edges in $\partial(V_1)$ between the
$i$th vertical segment and the $i+1$ vertical segment is at least
$|y_{i+1}-y_i|$. Thus, the number of horizontal edges in $\partial(V_1)$ is
at least
$$ |y_2-y_1|+\dotsb+|y_k-y_{k-1}| \ge Y-\eta. $$
As a result, we have at least
$$ \max \{ Y-\eta, a \} $$
horizontal edges in $\partial(V_1)$. Counting in the same way vertical edges,
we get
```
\begin{align*}
|\partial(V_1)|
&\ge \max \{ Y-\eta, a \} + \max \{ X-\xi, b \} \\
&\ge \frac12\,(X-\xi+a) + \frac12\,(Y-\eta+b).
\end{align*}
```

We now show that
$$ X-\xi+a\ge k. \tag{2} $$
Similarly, $Y-\eta+b\ge k$, and the two estimates readily yield the
assertion.

We observe that (2) is immediate if $X=k$ and $\xi=0$, and also if $X<k$ and
$\xi>0$ (when $a=k$). The two remaining cases can be dealt with as follows.

If $X<k$ and $\xi=0$, then $a$ is the number of those indices $i\in[1,k]$
with $x_i>0$. Therefore, in view of (1), we have $k^2/4\le aX$, whence
$$ X-\xi+a = X+a \ge k, $$
as wanted.

Finally, if $X=k$ and $\xi>0$, then $a$ is the number of those $i\in[1,k]$
with $x_i<k$. Hence, (1) gives
$$ k^2/4 \ge a\xi+(k-a)k = k^2 - a(k-\xi), $$
implying $a(k-\xi)>k^2/4$ and, as a result,
$$ X-\xi+a = a+(k-\xi) > k. $$