Here is a more detailed answer which somehow sums up the comments and remarks above.

Let $X$ be complex projective manifold and $[\alpha]\in H^{1,1}(X,\mathbb Z)$ be an integral $(1,1)$-Hodge class. The Lefschetz theorem on $(1,1)$-classes says that the first Chern class map $\operatorname{Pic}(X)\to H^{1,1}(X,\mathbb Z)$ is in fact surjective. 

This means that there exists on $X$ a line bundle $L\to X$ such that $c_1(L)=[\alpha]$.

Now, on a smooth projective manifold one always has $\operatorname{Pic}(X)\simeq\operatorname{Div}(X)/\sim$, where $\sim$ is linear equivalence of divisors. Thus, there exists an integral divisor $D=\sum a_j D_j$, $a_j\in\mathbb Z$ not necessarily positive, on $X$ such that the associated line bundle $\mathcal O_X(D)\simeq L$. In particular, $c_1(L)$ equals the cohomology class $[D]$ of $D$ (that is the Poincaré dual of the homology class of $D$). The integral (in general non-effective) cycle $D$ is what you are looking for.

Notice that the above isomorphism between the Picard group and the linearly equivalence classes of divisors can be rephrased saying that $L$ does always admit a (a priori) meromorphic section $s$: then set $D=\operatorname{div}(s)$.

Next, when can $[\alpha]$ be represented by an effective cycle (i.e. with all the $a_j\ge 0$)?. This is a much more subtle question.

Observe, first of all, that effectiveness is *not* in general a numerical property (i.e. it does not depend only on the first Chern class of the line bundle, or rather on how it intersects curves). To see this let $C$ be a smooth curve. Then, of course, two line bundles are numerically equivalent if and only if they have the same degree. If the genus $g$ of $C$ is greater than $1$, then the map
$$
\Phi\colon C\to\operatorname{Jac}^1(C)
$$ 
to the $g$-dimensional torus $\operatorname{Jac}^1(C)$ parametrizing degree $1$ line bundles on $C$ is an embedding. Its image corresponds to line bundles of the form $\mathcal O_C(p)$, for some $p\in C$, which are exactly the effective degree $1$ line bundles. Since $\dim\operatorname{Jac}^1(C)>\dim C$ by the hypothesis on the genus, you see that you can find a degree $1$ line bundle which is not effective (note also, in passing, that all line bundles of positive degree on a curve are ample, so that this example shows also that you can find non-effective ample line bundles -even if they eventually are, taking their tensor powers).

Of course, if $A\to X$ is a very ample or globally generated line bundle, then it is effective. Unfortunately, neither very ampleness nor globally generation are numerical properties (while ampleness, as well as nefness, bigness and pseudo-effectiveness are).

So, what you can do instead is to look at rational objects, that is to "tensor" the above $\mathbb Z$-modules by $\mathbb Q$. In this way, you can look at powers of a line bundle, hoping that it will become eventually effective and then "divide" by the same power to come back to the  starting object. 

In this way, if you know that $[\alpha]$ is big or ample (which, I repeat, depends only on $[\alpha]$), then you can conclude that it is $\mathbb Q$-effective: a high power $L^{\otimes m}$ in this case will have a global holomorphic section so that $L^{\otimes m}\simeq\mathcal O_X(D)$, with $D$ integral and effective, and thus $[\alpha]=[\frac 1m D]$ in rational cohomology.