= Solution
The joint <likelihood> is $\lambda^n e^{-\lambda T_n}\prod_i\mathbf1_{x_i>0}$, so the <factorization criterion for sufficiency> proves sufficiency. For any two positive sample vectors its likelihood ratio is $e^{-\lambda(T_n(x)-T_n(y))}$, independent of $\lambda$ exactly when their sums agree. The <likelihood-ratio criterion for minimal sufficiency> therefore makes $T_n$ minimal sufficient.
Its <gamma distribution> has density $\lambda^nt^{n-1}e^{-\lambda t}/\Gamma(n)$ on $t>0$. If $g(T_n)$ is integrable for every $\lambda$ and has expectation zero for every $\lambda$, then
$$
\int_0^\infty g(t)t^{n-1}e^{-\lambda t}\,dt=0\qquad(\lambda>0).
$$
By uniqueness of the <Laplace transform>, $g(t)t^{n-1}=0$ almost everywhere. One may first multiply by $e^{-\lambda_0t}$ to obtain an integrable function and use transform parameters $\lambda>\lambda_0$; this handles the transform's integrability hypothesis. Since the gamma densities are positive on $(0,\infty)$, $g(T_n)=0$ almost surely for every $\lambda$. \b[$T_n$ is complete as well as minimal sufficient.]
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