= Solution
For a dominated family of sample distributions, a statistic $T$ is a <sufficient statistic> precisely when the <probability mass function> or <density> factors as
$$
p_\theta(x)=g_\theta(T(x))h(x),
$$
where $h$ does not depend on the parameter. This is the <Fisher-Neyman factorization theorem>.
Here is the discrete proof in both directions. Suppose the factorization holds, and write $H(t)=\sum_{x:T(x)=t}h(x)$. Then $\mathbb P_\theta(T=t)=g_\theta(t)H(t)$. Whenever this probability is positive, $H(t)$ is finite and positive and
$$
\mathbb P_\theta(X=x\mid T=t)=\frac{h(x)}{H(t)},\qquad T(x)=t,
$$
is independent of $\theta$. Thus the <conditional distribution> of the whole sample given $T$ is parameter-free, which is the definition of a <sufficient statistic>.
Conversely, if $T$ is sufficient, let $q_t(x)$ denote this common conditional mass function on each possible fiber. Then
$$
p_\theta(x)=\mathbb P_\theta(T=T(x))q_{T(x)}(x).
$$
Take $g_\theta(t)=\mathbb P_\theta(T=t)$ and $h(x)=q_{T(x)}(x)$ to obtain the factorization. Fibers impossible for every parameter can be assigned $h=0$; conditional probabilities are required only for fibers with positive probability.
For the independent <Poisson> sample the joint mass function is
$$
p_\theta(x_1,\ldots,x_n)=\frac{e^{-n\theta}\theta^{\sum_i x_i}}{\prod_i x_i!},\qquad x_i\in\{0,1,2,\ldots\}.
$$
All parameter dependence is through
$$
\boxed{T=\sum_{i=1}^nX_i},
$$
so this is a one-dimensional <sufficient statistic>. The sample mean contains the same information, since $n$ is known.
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