Any non-singular projective variety over $\mathbb{C}$ is easily seen to be a smooth manifold. Presumably the same is not true for algebraic varieties – one would not expect varieties with singular points to have a smooth structure. But do there exist non-singular varieties that are not smooth manifolds?
Algebraic Geometry – Algebraic Varieties That Are Also Manifolds
ag.algebraic-geometry
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Here is a list biased towards what is remarkable in the complex case. (To the potential peeved real manifold: I love you too.) By "complex" I mean holomorphic manifolds and holomorphic maps; by "real" I mean $\mathcal{C}^{\infty}$ manifolds and $\mathcal{C}^{\infty}$ maps.
Consider a map $f$ between manifolds of equal dimension. In the complex case: if $f$ is injective then it is an isomorphism onto its image. In the real case, $x\mapsto x^3$ is not invertible.
Consider a holomorphic $f: U-K \rightarrow \mathbb{C}$, where $U\subset \mathbb{C}^n$ is open and $K$ is a compact s.t. $U-K$ is connected. When $n\geq 2$, $f$ extends to $U$. This so-called Hartogs phenomenon has no counterpart in the real case.
If a complex manifold is compact or is a bounded open subset of $\mathbb{C}^n$, then its group of automorphisms is a Lie group. In the smooth case it is always infinite dimensional.
The space of sections of a vector bundle over a compact complex manifold is finite dimensional. In the real case it is always infinite dimensional.
To expand on Charles Staats's excellent answer: few smooth atlases happen to be holomorphic, but even fewer diffeomorphisms happen to be holomorphic. Considering manifolds up to isomorphism, the net result is that many complex manifolds come in continuous families, whereas real manifolds rarely do (in dimension other than $4$: a compact topological manifold has at most finitely many smooth structures; $\mathbb{R}^n$ has exactly one).
On the theme of zero subsets (i.e., subsets defined locally by the vanishing of one or several functions):
One equation always defines a codimension one subset in the complex case, but {$x_1^2+\dots+x_n^2=0$} is reduced to one point in $\mathbb{R}^n$.
In the complex case, a zero subset isn't necessarily a submanifold, but is amenable to manifold theory by Hironaka desingularization. In the real case, any closed subset is a zero set.
The image of a proper map between two complex manifolds is a zero subset, so isn't too bad by the previous point. Such a direct image is hard to deal with in the real case.
Here are some facts.
If
$$ P(t) = a_0+a_1t+\cdots + a_{2k} t^{2k} $$
is a polynomial with nonnegative integral coefficients such that
$$ a_0=a_{2k}=1,\;\;a_j=a_{2k-j},\;\;\forall j $$
and $\newcommand{\bZ}{\mathbb{Z}}$
$$ a_k\in 2\bZ, $$
then there exists a smooth, compact, connected, oriented manifold $M$ of dimension $2k$ whose Poincare polynomial $P_M$ is the above polynomial $P$, i.e.,
$$b_j(M)=a_j,\;\;\forall j. $$
The manifold $M$ can be found by taking connected sums of products of spheres $S^{k_1}\times \cdots \times S^{k_m}$. The result is sharp in the following sense. There do not exist oriented smooth manifolds whose Poincare polynomials are
$$1 +t^6 +t^{12},\;\; 1+ t^{10}+ t^{20}. $$
This last fact was observed by Serre and follows from Hirzebruch's signature theorem.
Best Answer
Every non-singular algebraic variety over $\mathbb C$ is a smooth manifold. See for instance: http://en.wikipedia.org/wiki/Manifold under "Generalizations of Manifolds".
In fact, Arminius' suggested answer in the comments seems to give a proof of this fact, and I'll attempt to flesh it out a small amount. Every algebraic variety is locally a quasi-affine variety. So we may take an open cover $U_i$ of the variety, where each $U_i$ is a closed subset of an open subset of affine n-space. We may then check smoothness at each point of $U_i$ via the Jacobian criterion. The same procedure illustrates that each $U_i$ is a complex manifold. Since the gluing maps are algebraic, they are smooth, and hence our non-singular variety is also a smooth manifold.