Is there a topology such that $(\Bbb R, +, \mathcal T)$ is a compact Hausdorff topological group?

A good way to think about this is in terms of Pontryagin duality. Since we only care about the abelian group structure of $\mathbb{R}$, let's first get a nice characterization of this structure. As an abelian group, $\mathbb{R}$ is the unique $\mathbb{Q}$-vector space of its cardinality (up to isomorphism). An abelian group $A$ is a $\mathbb{Q}$-vector space iff for each nonzero $n\in \mathbb{Z}$, the multiplication by $n$ map $n:A\to A$ is an isomorphism.

Now the neat thing about this is that this condition is self-dual under Pontryagin duality. If $A$ is a locally compact abelian group, then the dual of the map $n:A\to A$ is just the map $n:\hat{A}\to\hat{A}$ on the dual group. So this says that a locally compact abelian group is a $\mathbb{Q}$-vector space iff its dual is a $\mathbb{Q}$-vector space.

In particular, let us use this to classify the compact abelian groups which are isomorphic (as groups) to $\mathbb{R}$. These are just the Pontryagin duals $\hat{V}$ of all $\mathbb{Q}$-vector spaces $V$ (with the discrete topology) for which $\hat{V}$ has cardinality $2^{\aleph_0}$. It is not hard to show that $\hat{\mathbb{Q}}$ has cardinality $2^{\aleph_0}$. If $V$ is a $\kappa$-dimensional $\mathbb{Q}$-vector space then $\hat{V}$ is a product of $\kappa$ copies of $\hat{\mathbb{Q}}$, which has cardinality $2^{\aleph_0\cdot\kappa}$.

So to sum up, there are indeed compact group topologies on $\mathbb{R}$. Up to continuous isomorphism, there is one such topology for each cardinal $\kappa$ such that $2^{\aleph_0\cdot\kappa}=2^{\aleph_0}$ (in particular, this includes all $\kappa$ such that $0<\kappa\leq\aleph_0$). The Pontryagin dual of this compact group is a $\mathbb{Q}$-vector space of dimension $\kappa$.

The case $\kappa=1$ gives $\hat{\mathbb{Q}}$, which is a solenoid. Explicitly, $\hat{\mathbb{Q}}$ is the inverse limit of the sequence $\dots\to S^1\stackrel{4}{\to}S^1\stackrel{3}{\to}S^1\stackrel{2}{\to}S^1$, since $\mathbb{Q}$ is the direct limit of the sequence $\mathbb{Z}\stackrel{2}{\to}\mathbb{Z}\stackrel{3}\to\mathbb{Z}\stackrel{4}{\to}\mathbb{Z}\to\dots$. For general $\kappa$, you just have a product of $\kappa$ copies of $\hat{\mathbb{Q}}$.