What is known about Lie groups with (strictly) positive curvature? If we consider $G$ a compact Lie group, there is a left invariant Riemannian metric whose the sectional curvature is nonnegative (see Milnors' paper). When can we find a left invariant metric that has positive sectional curvature?
The unique simply connected Lie group with a left-invariant metric that is positively curved is $SU(2)$. This is in Milnor's paper cited by @Igor Rivin.
Is $\mathrm{SU}(2)$ the only positively curved simply connected Lie group (not necessarily with an invariant metric)?
How could we distinguish the positively curved Lie groups?
Are there obstructions in the case of Lie groups? The fact that $G$ is a Lie group could imply the existence of some good invariant of positively curved manifolds?
Remark: I made many edits on the question because some ideas are clearer now with help of the time and the answers and comments posted here. I would like to observe that I started with a question with a pseudo-conjecture that the only positively curved Lie groups are $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ and asked for references.
 A: The standard reference is:
Curvatures of left invariant metrics on lie groups
by J Milnor - ‎1976 
In particular, a theorem of Wallace (mentioned in Milnor's paper) confirms your conjecture.
(PS: The paper seems to be available to all).
A: The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book.

Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive curvature.  Then $X$ has a zero.

Using this result, it's very easy to prove the following generalization of Wallach's theorem.

Theorem:  No compact Lie group $G$ of rank $2$ or higher admits a positively curved Riemannian metric which is invariant under left multiplication by any $T^2\subseteq G$.

Proof:  Suppose there is such a group $G$.  Because the $T^2$ action on $G$ by left multiplication is free, the action fields associated to it have no zeros.  On the other hand, because the $T^2$ is isometric, the action fields are are Killing fields.  By the above result, this implies the dimension of $G$ is odd.
Because the action of $S^1\times \{1\}\subseteq T^2$ on $G$ is isometric, the even dimensional manifold $G/(S^1\times \{1\})$ inherits a Riemannian metric for which the action by $\{1\}\times S^1$ is free and isometric.  Further, by O'Neill's formulas for a Riemannian submersion, the induced metric on $G/(S^1\times \{1\})$ is positively curved.  But then, just as above, the action field is non-zero and Killing, contradicting the quoted result above.  $\square$
Incidentally, without some kind of symmetry assumption, we can say basically nothing about the existence of positively curved metrics on compact Lie groups with finite $\pi_1$.  The issue that among closed simply connected manifolds, there is no known obstruction which distinguishes those that admits metrics of non-negative curvature from those which admit metrics of positive curvature.
