A question about Nash Equilibrium
From symmetry we can say that for each player $i$ in $I$, $q_r=q_k$ $\forall r,k \in I$. Similarly for $II$. Let this be $q_I$ for producers in $I$ and $q_{II}$ for $II$.
So profit function for individual $i$ in $I$:
\begin{align}
\max_{q_i} \pi_i&amp;=(200-(N-1)q_I-Nq_{II}-q_i))q_i-c_1q_i\\
\text{FOC: }\,\,\frac{\partial \pi_i}{\partial q_i}&amp;=200-(N-1)q_I-Nq_{II}-q_i-q_i-c_1=0 \\
\implies q_i^*(q_I,q_{II})&amp;=\frac{200-Nq_{II}-c_1-(N-1)q_I}{2}
\end{align}
Similarly, we will get for individual $j$ in $II$:
$$q_j^*(q_I,q_{II})=\frac{200-Nq_I-c_2 - (N-1)q_{II}}{2N}$$
At NE, we know that $q_i^*=q_I^*$ and $q_j^*=q_{II}$. Using this:
\begin{align}
q_I^*&amp;=\frac{200-Nq_{II}^*-c_1-(N-1)q_I^*}{2} \\
&amp;=\frac{200-Nq_{II}^*-c_1}{N+1}
\end{align}
And so, $$q_{II}^*=\frac{200-Nq_{I}^*-c_2}{N+1}$$
Solving for $q_I^*$:
\begin{align}
(N+1)q_I^*&amp;=200-N\frac{200-Nq_I^*-c_2}{N+1}-c_1 \\
\implies q_I^* &amp;=\frac{200-c_1+N(c_2-c_1)}{2N+1}
\end{align}
Similarly, $$q_{II}^* =\frac{200-c_2+N(c_1-c_2)}{2N+1}$$