<ol>
<li>The most obvious one is an <b>ideal single-photon source</b>
$$P_\text{id. s-p}(N=n) =
\begin{cases}
1 & n=1 \\
0 & \text{else.}
\end{cases}$$
Its shot noise is sub-Poissonian $\langle \Delta N^2 \rangle_\text{id. s-p} = \langle \left( N - \langle N \rangle \right)^2 \rangle_\text{id. s-p} = 0 < \langle N \rangle_\text{id. s-p} = 1$.</li>
<li>But also a <b>non-ideal single-photon source</b> is sub-Poissonian. For example, if it doesn't always generate a photon; may it be due to a finite excitation probability, quantum efficiency or collection efficiency. If the end-to-end probability to generate a single photon is $\eta$, the photon number distribution is
$$P_\text{imp. s-p}(N = n) =
\begin{cases}
\eta & n = 1 \\
1 - \eta & \text{else.}
\end{cases}
$$
Its variance $\langle \Delta N^2 \rangle_\text{imp. s-p} = \eta - \eta^2$ is still lower than the mean photon number $\langle N \rangle_\text{imp. s-p} = \eta$.</li>
<li>Due to the properties of a mathematical <a href="https://en.wikipedia.org/wiki/Convolution">convolution</a>, <b>all incoherent mixtures of a sub-Poissonian and a Poissonian distribution</b> are sub-Poissonian. "Incoherent mixture" means to simply add up the photons, without interference. Just like you add the results of two dice. For example, you could have an ideal single photon source emitting on top of Poissonian light. The resulting photon number distribution is then given by the convolution of the individual distributions
$$P_\text{tot}(N=n) = (P_\text{id. s-p} * P_\text{Poiss})(N=n).$$
The resulting photon number distribution has a mean value which is the sum of the mean values of the individual distributions $\langle N \rangle_\text{tot} = \langle N \rangle_\text{id. s-p} + \langle N \rangle_\text{Poiss}$. The variances also add up $\langle \Delta N^2 \rangle_\text{tot} = \langle \Delta N^2 \rangle_\text{id. s-p} + \langle \Delta N^2 \rangle_\text{Poiss}$. Since $\langle \Delta N^2 \rangle_\text{id. s-p} < \langle N \rangle_\text{id. s-p}$ and $\langle \Delta N^2 \rangle_\text{Poiss} = \langle N \rangle_\text{Poiss}$, we have $\langle \Delta N^2 \rangle_\text{tot} < \langle N \rangle_\text{tot}$.</li>
</ol>