= Solution
Use the curvature convention $\mathcal R$ of Solution 1, and put $T=\dot\gamma$. On continuous piecewise smooth perpendicular <vector fields> with zero endpoint values, the <Riemannian index form> is the symmetric bilinear form
$$
\boxed{I(V,W)=\int_0^L\bigl(\langle D_tV,D_tW\rangle-
\langle\mathcal R(V,T)T,W\rangle\bigr)dt.}
$$
A <Jacobi field> is a smooth field $J$ along $\gamma$ satisfying $D_t^2J+\mathcal R(J,T)T=0$. The point $\gamma(t_0)$ is a <conjugate point> to $\gamma(0)$ along this <geodesic> when there is a nonzero <Jacobi field> with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $\langle J,T\rangle$ is affine and vanishes at both endpoints.
The <conjugate-point criterion for the Riemannian index form>, stated without proof, is
$$
\boxed{\begin{aligned}
I(V,V)\ge0\text{ for all }V
&\iff\text{no conjugate time lies in }(0,L),\\
I(V,V)>0\text{ for every nonzero }V
&\iff\text{no conjugate time lies in }(0,L].
\end{aligned}}
$$
At a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing <Jacobi fields>. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.
Choose $n-1$ perpendicular orthonormal fields $E_1,\ldots,E_{n-1}$ by <parallel transport> along $\gamma$, and put $V_i(t)=\sin(\pi t/L)E_i(t)$. They are admissible and
$$
\sum_{i=1}^{n-1}I(V_i,V_i)
=\int_0^L\left((n-1)\frac{\pi^2}{L^2}\cos^2\frac{\pi t}{L}
-\sin^2\frac{\pi t}{L}\operatorname{Ric}(T,T)\right)dt.
$$
The tangent has unit length. The assumed <Ricci curvature> bound and the two elementary sine-square and cosine-square integrals give the <sine index-form bound for positive Ricci curvature>:
$$
\boxed{\sum_{i=1}^{n-1}I(V_i,V_i)
\le\frac{n-1}{2}\left(\frac{\pi^2}{L}-\kappa L\right)<0
\quad\text{when }L>\frac\pi{\sqrt\kappa}.}
$$
Since $n\ge2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.
If the <Riemannian manifold> is complete, the <Hopf-Rinow theorem> supplies a <minimizing geodesic> between any two points. Its <Riemannian index form> is nonnegative: any smooth fixed-endpoint variation has nonnegative <second variation of geodesic energy>, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the <Cauchy-Schwarz inequality>. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi/\sqrt\kappa$. Taking the supremum over pairs yields the diameter conclusion of the <Bonnet-Myers theorem>:
$$
\boxed{\operatorname{diam}(M,d_g)\le\frac\pi{\sqrt\kappa}.}
$$
The completeness assumption is essential. An explicit <incomplete positively curved strip with infinite diameter> is
$$
M=(-\pi/4,\pi/4)\times\mathbb R,\qquad g=du^2+\cos^2u\,dv^2.
$$
The map $(u,v)\mapsto(\cos u\cos v,\cos u\sin v,\sin u)$ is a <local isometry> to the unit <sphere>, so $K=1$ and, in dimension two, $\operatorname{Ric}=g$. Thus the required <Ricci curvature> bound holds with $n=2$, $\kappa=1$. The unit-speed meridian $t\mapsto(t,0)$ reaches the missing boundary $u=\pi/4$ in finite time, so the strip is <geodesically incomplete>. On the other hand every <curve> joining $(0,0)$ to $(0,A)$ has
$$
L_g(c)\ge\frac1{\sqrt2}\int|v'(t)|\,dt\ge\frac{|A|}{\sqrt2},
$$
since $\cos u\ge1/\sqrt2$ throughout the strip. Therefore \b[its intrinsic diameter is infinite], in particular larger than $\pi$. This is the unwrapped strip with $v\in\mathbb R$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.
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