= Solution
By the <Hodge decomposition theorem>, the first <Betti number> is $b_1(M)=\dim\mathcal H^1(M)$. Part (b) makes every element of this space a <parallel differential form>. For any fixed $p\in M$, consider the linear evaluation map
$$
\operatorname{ev}_p:\mathcal H^1(M)\longrightarrow T_p^*M,\qquad\omega\longmapsto\omega_p.
$$
It is injective. A parallel form with $\omega_p=0$ remains zero under <parallel transport> along any path from $p$; <connectedness> of the <smooth manifold> makes every point reachable by such a path. Equivalently, the derivative of the squared norm of a parallel form is zero in every direction, so its norm is constant on the connected manifold. Either argument gives $\omega=0$ globally. Since $\dim T_p^*M=n$, the <dimension bound for harmonic one-forms under nonnegative Ricci curvature> gives
$$
\boxed{b_1(M)\le n.}
$$
For an explicit obstruction in dimension three, take $N=\Sigma_2\times S^1$, where $\Sigma_2$ is a closed orientable surface of genus two. It is a compact connected smooth three-manifold. A cell decomposition of $\Sigma_2$ has one vertex, four one-cells and one two-cell attached by a product of commutators. Its cellular boundary into the one-cells is zero, because each generator has total exponent zero in that attaching word; the boundary of every one-cell is also zero. Therefore $H^1(\Sigma_2;\mathbb R)=\mathbb R^4$. The <Künneth theorem> in degree one and $H^1(S^1;\mathbb R)=\mathbb R$ give
$$
H^1(N;\mathbb R)\cong H^1(\Sigma_2;\mathbb R)\oplus H^1(S^1;\mathbb R)\cong\mathbb R^5.
$$
Consequently
$$
\boxed{\dim N=3,\qquad b_1(N)=5>3,\qquad
N\text{ admits no metric with }\operatorname{Ric}\ge0.}
$$
This is the <Betti-number obstruction to nonnegative Ricci curvature>. It rules out every <Riemannian metric> on that underlying manifold, not merely a particular product metric.
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