In An Introduction to Tensor Calculus and Relativity, Lawden provides the definition of mass by considering a collision between two massive particles $p$ and $q$ with masses $m_p$ and $m_q$, respectively. The particles' are given velocities $\mathbf u_p,\mathbf u_q$, respectively, prior to collision, and $\mathbf v_p,\mathbf v_q$, after. It is has been shown experimentally that, in general, $m_p\mathbf u_p+m_q\mathbf u_q=m_p\mathbf v_p+m_q\mathbf v_q$ and from this it follows, or so Lawden claims, that
$$\frac{m_p}{m_q}=\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert}.$$
If $q$ is then chosen to have unit mass, we define the mass of any particle by observing a collision between it and $q$.
However, this equation holds only in the case that $\frac{m_p}{m_q}$ is positive, in general we should only have
$$\left\lVert\frac{m_p}{m_q}(\mathbf u_p-\mathbf v_p)\right\rVert=\lVert\mathbf v_q-\mathbf u_q\rVert,$$
which gives us
$$\left\lvert\frac{m_p}{m_q}\right\rvert=\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert}.$$
So if we want to define mass this way, we ought to have
$$\frac{m_p}{m_q}=\operatorname{sgn}\left(\frac{m_q}{m_p}\right)\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert},$$
which is positive if $m_p,m_q$ have the same sign, and negative otherwise. Since the masses aren't defined a priori, the sign should be a free parameter unless there's some conserved relation between the velocities which necessitates that mass is always positive.
Edit/Remark
It is readily apparent that classical laws of motion are invariant under simultaneous inversion of mass and parity:
$$\mathbf F=m\mathbf a=(-1)^2m\mathbf a=(-m)(-\mathbf a)=-m\frac{\partial^2}{\partial t^2}(-\mathbf x)$$
$$\mathbf p=m\mathbf v=(-1)^2m\mathbf v=(-m)(-\mathbf v)=-m\frac{\partial}{\partial t}(-\mathbf x)$$
$$E^2=p^2c^2+m_0^2c^4=p^2c^2+(-m_0)^2c^2$$
$$\mathbf F=G\frac{m_1m_2}{\lVert\mathbf r_1-\mathbf r_2\rVert^3}\lVert\mathbf r_1-\mathbf r_2\rVert=G\frac{(-1)^2m_1m_2}{\lVert-(\mathbf r_1-\mathbf r_2)\rVert^3}\lVert-(\mathbf r_1-\mathbf r_2)\rVert=G\frac{(-m_1)(-m_2)}{\lVert(-\mathbf r_1)-(-\mathbf r_2))\rVert^3}\lVert-((-\mathbf r_1)-(-\mathbf r_2))\rVert$$
I haven't checked the laws for electromagnetism yet, but I would guess that the same holds with, at most, the inversion of charge. In this sense, the sign of mass is a matter of convention, one which reflects the sign convention for position. However, this is not what I am interested in. What I want to know is whether or not it is necessary that all masses have the same sign as a consequence of some conserved relation which would not apply otherwise.