From the 2nd Newton's law one can get the following equations:
$$
\left\{
\begin{aligned}
\frac{d v_x}{d t} + \frac{F_a(v)}{m v} v_x & = 0 \\
\frac{d v_y}{d t} + \frac{F_a(v)}{m v} v_y & = -g
\end{aligned}
\right. \qquad (1)
$$
where
$v = \sqrt{v_x^2 + v_y^2}$,
$m$ is the mass of the bullet,
$\vec{F}_a$ is the force of the air resistance,
$\vec{g}$ is the free fall acceleration.
If we assume the air resistance force to fit Stoke's law:
$$
\vec{F}_a = - 6 \pi \mu R \vec{v}
$$
where
$\mu$ is the viscosity,
$R$ is the radius of the ball.
Then system (1) becomes:
$$
\left\{
\begin{aligned}
\frac{d v_x}{d t} + \frac{6 \pi \mu R}{m} v_x & = 0 \\
\frac{d v_y}{d t} + \frac{6 \pi \mu R}{m} v_y & = -g
\end{aligned}
\right. \qquad (2)
$$
This system can be solved analytically. You can find the solution and investigate it's dependence on the parameters of the ball and the angle.
The coefficient representing air resistance is proportional to $R/m$ rate. If the material is fixed the mass is proportional to $R^3$ and then this coefficient is proportional to $R^{-2}$. This means that the ball should not be too small. Indeed very small particles (dust) loose their velocity very fast.
If the throwing force (initial momentum) is fixed big mass leads to low initial velocity. So the ball should not be too large.
There is an optimal value of the mass and it depends on the angle. So you have to optimize both mass and angle simultaneously.
We can not neglect air resistance because it will lead to zero optimal mass.
Edit:
Stoke's law does not work for high Reynolds number which can result from high inital velocity and/or large size of the ball. If we use drag equation
$$
F_a(v) = \frac{1}{2} \rho v^2 C_D \pi R^2
$$
we will get a system of nonlinear ODEs. That system probably can not be solved analytically.
In this case air resistance will be proportional to $R^{-1}$ and all general conclusions will be the same as for Stoke's law, i.e. both huge and microscopic balls will not fly very far.
