The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the socalled formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?

I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of DeligneGriffithsMorganSullivan on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers. I am not totally sure, but I believe that the definitions are as follows:
One of the things I'm not sure about is whether in the definition we should require $H^\ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^\ast(X)$ is always (graded)commutative. The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^\infty$) differential forms on X with the standard de Rham differential is a formal dg algebra. The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal. Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally. In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of Ainfinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasiisomorphic to $H^\ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (KontsevichSoibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in this paper), and you can find it in the operads literature as the "transfer theorem") that you can put an Ainfinity structure on $H^\ast(A,d)$ which makes $A$ and $H^\ast(A,d)$ quasiisomorphic as Ainfinity algebras. The Ainfinity structure manifests itself as a series of $n$ary products satisfying various compatibilities. Intuitively at least, these $n$ary products should be thought of as being analogous to Massey products in topology. So $H^\ast(A,d)$ with this Ainfinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasiisomorphic, as an Ainfinity algebra, to $H^\ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^\ast(A,d)$. *Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the Ainfinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the Ainfinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of Ainfinity structure. From the paper:
and also
(Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.) 


Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :) Here it is: let $X$ be a simply connected differentiable manifold. Rational homotopy theory tells us that the rational homotopy type of $X$ (that is, its homotopy type modulo torsion) is contained in its minimal model, $M_X$, which is a commutative differential graded (cdg) algebra. By definition, this means that you have a quasiisomorphism (quis, a morphism of cdg algebras inducing an isomorphism in cohomology) $$ M_X \longrightarrow \Omega^*(X) \ . $$ Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the minimality of $M_X$ means that, in a certain, but precise, sense, it is the smallest cdg algebra for which such a quis exists. The fact that $M_X$ contains the rational homotopy type of $X$ implies, for instance, that you can obtain the ranks of the homotopy groups of $X$ from it:
Nice, isn't it? :) The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you. Almost by definition, $X$ is a formal space if there exists two quis $$ \Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q}) $$ Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra $H^*(X; \mathbb{Q})$, which is nicer (smaller, more computable) than $\Omega^*(X)$. And the final point is that there are plenty of examples of spaces which are known to be formal. (Final remark: Actually, you'd have to put $A_{PL}^*(X;\mathbb{Q})$ instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.) 


Maybe you could take a look at Y. Félix, J. Oprea, D. Tanré; Algebraic models in Geometry, Oxford Graduate Text in Math. 17 (2008) where they talk about formality in the context of rational homotopy theory, RHT, (for instance, in sections 2.7 and 3.1.4). Also the more classical, but excellent little book D. Lehmann; Théorie homotopique des formes différentielles, Astérisque 45 is worth reading (section V.9). As for formality in the context of operads, allow me a little selfpromotion :) : F. Guillén, V. Navarro, P. Pascual, Agustí Roig, Moduli spaces and formal operads; Duke Math. J. 129, 2 (2005). In this work, we translate some classical results concerning formality in RHT to chain operads. For instance, the DeligneGriffithsMorganSullivan theorem about formality of Kähler manifolds, formality's independence of the ground field... And extend them also to modular operads. 


Formal can mean slightly different things in different contexts. A commutative differential graded algebra (CDGA) is formal if it is quasiisomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the HalperinStasheff paper). To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasiisormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA. If A is a CDGA which is quasiisomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology. I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more strings of quasiisomorphisms between them as DGAs then as CDGAs. I believe it is unknown if two CDGA that are quasiisomorphic as DGA have to be quasiisomorphic as CDGA. 

