This is probably simple. Is there a finite non-commutative ring $R$ with identity in which all of its ideals are two-sided !?



non-commutative finite rings
A "natural" example is given by the group ring $\mathbb{F}\_2[Q]$ of the Quaternion group of order 8. 


For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in [this paper](http://www.ams.org/journals/proc/1979-076-02/S0002-9939-1979-0537074-2/S0002-9939-1979-0537074-2.pdf). 


Let $i:\mathbb{F}\_2[Q] \to \mathbb{F}\_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal. 


Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done. 


**Added:** The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}\_2[Q]$ can be embedded into the matrix ring $M\_4(\mathbb{F}\_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a faithful representation $Q\hookrightarrow GL(4,2)$ is given by 
$$x \mapsto 
\begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad
y \mapsto 
\begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$
Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D\_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}\_2$. 


