Assuming the metric convention is $(-,+,+,+)$, the action of a relativistic point particle with worldline $X^\mu(s)$ is
$$ S = -m \int ds \sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }. $$
To make this an integral over spacetime, we can insert an integration over a delta function
$$ S = \int d^4 x \left[ -m \int ds \delta^4(x - X(s) ) \sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu } \right]. $$
Now, the Hilbert stress-energy tensor is defined as
$$ T_{\mu\nu} = \frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu} }$$
Which in this case becomes
$$ T_{\mu\nu} = -m \int ds \frac{\delta^4(x - X(s))}{\sqrt{-g}} \frac{\dot{X}_\mu \dot{X}_\nu }{\sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }}.$$
This has the wrong sign, however. Choosing for the parameter $s$ the coordinate time $t$, we write $X^\mu = (t, \vec{X}(t))$ thus $\dot{X}^\mu = v^\mu= (1, \vec{v})$, with $\vec{v}$ the coordinate velocity. This allows one to perform the integration, leaving
$$ T_{\mu\nu}(t,\vec{x}) = -m \int dt' \frac{\delta(t - t')\delta^3(\vec{x}-\vec{X}(t'))}{\sqrt{-g}} \frac{\dot{X}_\mu \dot{X}_\nu }{\sqrt{-g^{\mu\nu} \dot{X}_\mu \dot{X}_\nu }} = -m \frac{\delta^3(\vec{x}-\vec{X})}{\sqrt{-g}} \gamma v_\mu v_\nu. $$
This implies $T_{00} \sim -\gamma m$ is negative, which cannot be right. What went wrong?