= Solution
The null model has only an intercept:
$$
Y_i\overset{\mathrm{iid}}\sim\operatorname{Poisson}(\mu),\qquad\log\mu=\beta_0.
$$
For $\bar Y>0$, maximizing its <log-likelihood> gives $\widehat\mu=\bar Y$ and $\widehat\beta_0=\log\bar Y$. The saturated <Poisson regression> gives each observation its own fitted mean $\widehat\mu_i=Y_i$, allowing zero as a limiting fitted mean. By the definition of <Poisson deviance>,
$$
D_0=2\sum_{i=1}^{107}\left\{Y_i\log\frac{Y_i}{\bar Y}-(Y_i-\bar Y)\right\}.
$$
Since $\sum_i(Y_i-\bar Y)=0$,
$$
\boxed{D_0=2\sum_{i=1}^{107}Y_i\log(Y_i/\bar Y).}
$$
Use the continuous convention $0\log(0/\bar Y)=0$. If all counts are zero, both the saturated and null likelihood maxima are attained as limits at zero mean and the deviance is zero.
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