I'm going to make an attempt to explain this from scratch, so here goes (WARNING: this might get complicated):
After symmetry breaking, the mass term for the weak field in the standard model Lagrangian becomes $\mathcal{L}{W, mass} = -\frac{1}{2\lambda}\left(\mu g_2\right)^2W^{-\nu}W^+{\nu} = m^2_WW^{-\nu}W^+_{\nu}$. For the Weak bosons to be massless requires $\mu g_2 = 0$. Therefore there are 2 ways to go about this:
1. $\mu = 0$
2. $g_2 = 0$
So what are these terms? $\mu = \frac{1}{\sqrt{2}}m_H$, the mass of the Higgs. $\lvert e\rvert = g_2\sin\theta_W$ is the (modulus of the ) charge of an electron. The mass of the Z boson is $m_Z = \frac{m_W}{\cos\theta_W}$
Considering the two cases separately:
1: Setting $\mu = 0$:
Both the Higgs and the Z boson also become massless. The Higgs mechanism no longer exists and electroweak symmetry breaking never occurs. The Higgs potential becomes $V = -\mu^2\phi^\cdot\phi + \frac{1}{2}\lambda\left(\phi^\cdot\phi\right)^2 = \frac{1}{2}\lambda\left(\phi^*\cdot\phi\right)^2$ which has a minimum point at $0$ (i.e. $\left\langle\phi_0\right\rangle_{min} = \left\langle\phi_1\right\rangle_{min} = v = 0$). Perturbing about this point with $\phi_0 = \frac{H}{\sqrt{2}}$ gives $V = \frac{\lambda H^4}{8}$
This also causes the mass of all elementary fermions to be $0$, or, the mass of all the elementary particles to be discovered so far is $0$. 298
As a result of this, the mass of the neutron and the proton are equal. Not that it matters, because current understanding has it that the universe would consist of nothing but plasma as a result of constant pair creation and annihilation of, well, all the fundamental particles. That's assuming that the fundamental particles would even be the same.
2: Setting $g_2 = 0$:
The first result is that nothing has electric charge. The concept of an electric charge ceases to exist. The gauge transformation for the gauge bosons becomes is $D_{\nu} = \partial_{\nu} + \frac{i}{2}Ig_1B_{\nu}$ where $I$ is the identity matrix. This gives the single field, $B_{\nu}$. In this case, $v = \frac{\mu}{\sqrt{\lambda}}$ as usual and the mass term becomes $\mathcal{L}{B, mass} = -\frac{1}{8}v^2g^2_1B{\nu}B^{\nu} = -\frac{1}{2}m_B^2B_{\nu}B^{\nu}$, giving a mass of $m_B = \frac{vg_1}{2}$. However, the three other gauge fields don't just vanish - they still exist, but are massless. These fields are not the same fields as the photon and W boson fields. Nevertheless, we have 3 massless fields and another with mass. From your comment saying that you want the Z boson to have mass, this is therefore the condition closest to what you want.
This is surprisingly simple: the <PERSON> term in the Standard Model Lagrangian consists of 3 massless gauge fields and 1 massive one, as opposed to the relative mess that results from the electroweak symmetry breaking of our own universe. The symmetry breaking still occurs, so I'll assume that the fundamental particles are still the same, although the other way round - the SU(2) symmetry is unbroken and the U(1) symmetry is broken.
However, there is a difference: the coupling constant of the 3 massless fields, $g_2 = 0$ and so they just don't interact with anything.
Looking at the leptons: we have 3 generations of $l = \begin{pmatrix} \nu_e \ e \end{pmatrix}$ and $\bar{e}$. 800