Does the sign of curvature remain constant under the sum of their metrics? Does the sign of curvature remain constant under the sum of their metrics over a fixed manifold? e.g. If $(M,g_1)$ and $(M,g_2)$ are closed Riemannian manifolds of positive curvature then $(M,g_1+g_2)$ is also of positive curvature?
Update: Q2: What if we weaken the problem by allowing manifolds having boundary?
 A: Edit  As C.F.G. noted, I had originally answered different but related question.  So, let me answer the actual question asked with a No: the sum of two positively curved metrics can have negative curvature.  I'll construct an example below.
To do so, consider $S^3$ with a metric $g_\lambda$ making it a Berger Sphere.  Recall that this is obtained by beginning with the usual round metric $g_0$ and scaling it in the direction of the Hopf fibers.  More precisely, given any vectors $v,w\in T_p S^3$, we define $g_\lambda(v,w) = g_0(v,w) + \lambda g_0(v',w')$ where $v'$ is the $g_0$-projection of $v$ to the tangent space of the Hopf circle through $p$, and likewise for $w'$.
It is known that the resulting metric $g_\lambda$ is positive curved iff $-1<\lambda < 1/3$.  (If $\lambda \leq -1$, then $g_\lambda(v,v) \leq 0$ for all $v$ tangent to a Hopf circle, so we don't even get a metric in this case).
Pick any $\lambda$ strictly between $0$ and $1/3$.
Now, consider the metric $g:=g_{\lambda} + t g_0$ on $S^3$.  For any $t > 0$, this is a sum of positively curved metrics.  However, I claim that for any $t$ with $0< t < 3 \lambda -1$, that $g$ has negative curvature planes.
To see this, note that \begin{align*} g(v,w) &= g_\lambda(v,w) + t g_0(v,w)\\ &= (1+t) g_0(v,w) + \lambda g_0(v',w'),\end{align*} so $$\frac{1}{1+t} g(v,w) = g_0(v,w) + \frac{\lambda}{1+t} g_0(v',w')$$ is a Berger metric $g_{\lambda/(1+t)}$.  Since scaling doesn't change the sign of curvature, it follows that if we can show that $\lambda/(1+t) > \frac{1}{3}$, we are done.  Since $t< 3 \lambda - 1$, we find $\lambda/(1+t) > \lambda/(1 + 3\lambda - 1) = \frac{1}{3}.$
End edit - the old answer is left below
Proposition:  There is a positive curved metric $g_b$
No, the sum of metrics of fixed signed curvature is, in general, not of the same sign.
Proposition:  If the sum of any two positively curved metrics is positively curved, then every closed manifold of dimension $2$ or more admits a metric of positive curvature.
Proof:  First, if $g$ is a metric with positive curvature, then $\lambda g$ is for any positive $\lambda$.  Also, an easy induction argument shows that the hypothesis imply that any finite sum of positively curved metrics is positively curved.
Now, given any closed manifold $M^n$, compactness allows us to cover $M$ by a finite collection of open sets $U_1,..., U_k$, each diffeomorphic to $\mathbb{R}^n$ with the property that their closures $\overline{U_i}$ are contained in open sets $V_i$ which are also diffeomorphic to $\mathbb{R}^n$.  Let $\lambda_i$ be a partition of unity supported on $\overline{U}_i$.
Since $n\geq 2$, each $V_i$ admits a metric $g_i$ which is positively curved.  For example view $\mathbb{R}^n$ as the complement of a point in $S^n$ then pull the induced metric back to $g_i$.
Now the metic $g = \sum \lambda_i g_i|_{U_i}$ is positively curved by the first paragraph.  $\square$
Of course, there are obstructions to admitting a positively curved metric.  So, a consequence of this is that the sum of positively curved metrics need not be positively curved.
