I claim that every finite $T_{0}$ critical space is isomorphic to $(n,\tau_{n})$.

Suppose that $X$ is a finite $T_{0}$-space. Let $\leq$ be the specialization ordering on $X$. Then $U\subseteq X$ is open if and only if $U$ is upwards closed. Then each finite $T_{0}$-space has an injective selector map, namely $x\mapsto{\uparrow}x$ is a neighborhood selector map.

Now suppose that $X$ is not linearly ordered. Then there are two elements $r,s\in X$ with $r\not\leq s$ and $s\not\leq r$. Then ${\uparrow}r\cup{\uparrow}s$ has no least element, so ${\uparrow}r\cup{\uparrow}s\neq{\uparrow}x$ whenever $x\in X$. Therefore $|X|<|\mathcal{T}\setminus\{\emptyset\}|$, so there is no surjective neighborhood selector map.

Therefore if $X$ is a finite critical $T_{0}$-space, then the specialization ordering on $X$ is a linear ordering and hence isomorphic to $(n,\tau_{n})$.

$\textbf{added 3/14/2015}$

I claim that the specialization ordering on any critical $T_{0}$-space must be well-founded. Suppose to the contrary that $(X,\mathcal{T})$ is a critical $T_{0}$-topological space whose specialization ordering is not well founded. Then there is some injective neighborhood selection map $j:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$. However, since $X$ is not well-founded, there is a descending sequence $x_{0}>x_{1}>x_{2}>x_{3}>...$ Since $j(x_{n+1})$ is an upwards closed set containing $x_{n+1}$, we have $x_{n}\in j(x_{n+1})$ as well.

Now define a mapping $k:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$ by letting $k(x_{n})=j(x_{n+1})$ for all $n\in\omega$ and where $k(x)=j(x)$ whenever $x\in X\setminus\{x_{n}|n\in\omega\}$. Then $k$ is an injective neighborhood selector map which is not surjective onto $\mathcal{T}\setminus\{\emptyset\}$. We conclude that every critical $T_{0}$-topological space has a well-founded specialization ordering.

I however claim that not every $T_{0}$-critical space is isomorphic to $(\alpha,\tau_{\alpha})$. Let $X=(\omega+1)\cup\{0'\}$ where $0'$ takes the place of an extra zero. Partially order $X$ by letting $\omega+1$ have its original partial ordering and where $0'\leq n$ whenever $n\neq 0$ but where $0\not\leq 0',0'\not\leq 0$. Give $X$ the topology $\mathcal{T}$ where a set $U\subseteq X$ is declared to be open if $U$ is upwards closed and where if $\omega\in U$ then $n\in U$ for some $n\in\omega$. Then $\leq$ is the specialization ordering for the topology $\mathcal{T}$.

I first claim that there is an injective neighborhood selector function. Simply let
$j(0)=X\setminus\{0'\},j(0')=X\setminus\{0\},j(\omega)=X$ and
$j(n)=\uparrow n$ for all $n\in\omega\setminus\{0\}$.

However, I claim that every injective neighborhood selector function is bijective. Suppose that $j:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$ is a neighborhood selector function. Suppose that $j(\omega)\in X,X\setminus\{0\},X\setminus\{0'\}$. Then we have function $j[\{\omega,0,0'\}]\subseteq\{X,X\setminus\{0\},X\setminus\{0'\}\}$. Therefore $j$ maps $\{\omega,0,0'\}$ onto $\{X,X\setminus\{0\},X\setminus\{0'\}\}$. Therefore one can prove by induction that $j(n)=\uparrow n$ for all $n$.

Now assume that $j(\omega)=\uparrow n$ for some $n>0$. Then $j[\{\omega,0,...,n-1,0'\}]\subseteq\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}$.

However, since $|\{\omega,0,...,n-1,0'\}|=n+2$ and
$|\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}|=n+2$, we have
$j[\{\omega,0,...,n-1,0'\}]=\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}$. By induction, we can conclude $j(m)=\uparrow m$ for each $m\in\omega$ with $m>n$. In this case, the function $j$ will turn out to be surjective.

We conclude that the space $X$ is critical, but $X$ is not isomorphic to any $(\alpha,\tau_{\alpha})$.