Does mass bend spacetime always with a positive curvature? In cosmological models the curvature of spacetime at large scales is constant, and the sign does not change with time (correct me if I am wrong). But what happens at non-cosmological scales? let us say two masses orbiting, or more complicated systems, is the curvature induced by these masses always positive because gravity is attractive? (I am not sure if this connection is accurate). Would the hypothetical matter with negative energy (the one require to make wormholes) create a curvature of negative sign and also a repulsive force?
 A: This question can be answered by studying the Raychaudhuri equation and energy conditions in General Relativity.
The problem we're trying to answer essentially is

Which conditions are necessary for gravity to be attractive, i.e., for geodesics to get closer as they evolve with time?

To answer this, let us consider the Raychaudhuri equation. This is an equation in General Relativity that tells us about the evolution of the expansion of a family of geodesics, i.e., it tells us how much they are getting apart from each other. We can write it as (see Wald, Eq. (9.2.11))
$$\frac{\mathrm{d}\theta}{\mathrm{d}\tau} = - \frac{1}{3} \theta^2 - \sigma_{ab}\sigma^{ab} + \omega_{ab}\omega^{ab} - R_{cd}u^c u^d,$$
where $u^a$ is the four-velocity of the observers, $R_{ab}$ is the Ricci tensor, $\sigma_{ab}$ is a tensor known as the shear (it measures the shear on the family of geodesics) and $\omega_{ab}$ is the twist (how much the geodesics rotate).
To keep us dealing with the simplest of cases, I'll assume a priori that there is no shear or twist in the family of geodesics (this is true for comoving observers in FLRW, for example). Then the Raychaudhuri equation becomes
$$\frac{\mathrm{d}\theta}{\mathrm{d}\tau} = - \frac{1}{3} \theta^2 - R_{cd}u^c u^d.$$
$\theta$ is the expansion, so it measures how geodesics are getting far from each other. To get repulsive gravity, we need to have positive $\theta$ and positive $\frac{\mathrm{d}\theta}{\mathrm{d}\tau}$ (or the geodesics will get together again). From the Raychaudhuri equation, we see that a necessary condition to that will be that $R_{cd}u^c u^d < - \frac{1}{3} \theta^2 < 0$.
In technical terms, this means we need a violation of the strong energy condition (SEC). This is a requirement on the stress-energy tensor that is respected by much of the matter that we know of, but violated by some (for example, the cosmological constant). For a perfect fluid, it is equivalent to requiring that $\rho + p \geq 0$ and $\rho + 3 p \geq 0$. Notice it does not require $\rho > 0$, which means you can violate SEC without the need for matter with negative energy. Furthermore, having negative energy does not imply you will violate SEC, and hence it does not imply repulsive gravity.
A different way of viewing this is by using the FLRW solution (I'll omit the cosmological constant, so it becomes just a toy model to study solutions with perfect fluids). The Friedmann equations imply that the scale factor respects
$$\frac{\ddot{a}}{a} = - \frac{4\pi}{3} (\rho + 3p).$$
Hence, to have repulsive gravity ($\ddot{a} > 0$), we just need $\rho + 3p < 0$, which is precisely violating SEC. $\rho < 0$ alone is not sufficient if the pressure is too large, and it is not necessary if the pressure is too negative.
Conclusions
Gravity needs not to be attractive, that is a of the matter content of the theory. In principle, General Relativity does not forbid you from having repulsive gravity, as long as you give it the right matter content. This is just a consequence of the fact that you can make any metric solve the Einstein Equations by defining the stress-energy tensor by $T_{ab} = \frac{1}{8\pi}G_{ab}$. However, to have such matter around is a completely different story. As far as I know, the only known form of classical matter that violate SEC is the cosmological constant/dark energy. Quantum matter violates all usual energy conditions, so things get way more complicated. See, e.g., arXiv: 1208.5399 [gr-qc] for more on those aspects.

Ricci Scalar
To see whether the sign of the Ricci scalar is relevant, let us consider the Ricci scalar for FLRW. As quoted in Wikipedia, it is given by
$$R = 6 \left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{k}{a^2}\right).$$
Repulsive gravity would be $\ddot{a} > 0$, while attractive gravity would be $\ddot{a} < 0$. We see that these signs are not equivalent to those of the Ricci scalar. Hence, we can have repulsive gravity regardless of the sign of the Ricci scalar and we can have attractive gravity regardless of the sign of the Ricci scalar.
Let us consider it in even more generality. From the Einstein Equations,
$$R_{ab} - \frac{1}{2} R g_{ab} = 8 \pi T_{ab},$$
one can find that
$$R = - 8 \pi T_{ab}g^{ab}.$$
Suppose now an arbitrary object made of a perfect fluid. This doesn't need to be a cosmological model: one could model a star like this, for example.  Then one has $T_{ab}g^{ab} = - \rho + 3 p$. Hence, for any solution of the Einstein Equations with a perfect fluid,
$$R = 8 \pi (\rho - 3p),$$
meaning the sign of the Ricci scalar can be positive, vanishing, or negative depending on the particular equation of state.
A: In theory, that could be true as adding positive energy increases the curvature. Look at the Casimir effect for the nearest way negative energy affects repulsion. It shows that, if negative energy exists, it can cause the repulsion of ordinary matter.
Wavelengths between two objects are limited when placed in a space that has zero energy density. It has fewer wavelengths than in its surroundings. This means the energy density will go lower than zero and that is negative energy.
