**Disclaimer.** It turns out that as pointed out by user @Jason Gaitonde, the idea I  presented at the end of my question eventually solves my problem with the right choise of $N_1$, namely $N_1 = C \log N$ for sufficiently large positive constant $C$. In this post, I'll fill in the details. I'd be grateful if someone could kindly check the math. Thanks in advance.

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>**Claim.** *Take $N_1 = 2C \log N$, $k = N/N_1$, where $C$ is a sufficiently large positive constant. For large $N$, it holds w.p $1-1/N^{2C-1} =1-o(1)$ that*
$$
Z(\mathcal T) \gtrsim ((1-\theta)C \log N)^{N/(2C\log N)},
$$

*Proof.* First observe that, in the definision of $S_i$, we can further write $z_t = 2 b_t - 1$, where $b_t$ is Bernoulli with parameter $p=1-\theta/2 \in [1/2,1]$. Thus, $S_i = \sum_{t \in G_i} (2b_t - 1) = 2 B_i - N_1$, where $B_i := \sum_{t \in G_i} b_t \sim \mathrm{Bin}(N_1,p)$. By well-known concentration results, 
\begin{eqnarray}
\begin{split}
\mathbb P(B_i \ge (1+t)N_1 p) &\le e^{-\frac{t^2p^2N_1}{2+t}},\text{ for all }t > 0,\\
\mathbb P(B_i \le (1-t)N_1 p) &\le e^{-\frac{t^2p^2N_1}{2}},\text{ for all }0 < t < 1.
\end{split}
\end{eqnarray}
We deduce that,
\begin{eqnarray}
\begin{split}
\mathbb P(S_i \ge (2p(1+t) - 1)N_1) &\le e^{-\frac{t^2p^2N_1}{2+t}},\text{ for all }t > 0,\\
\mathbb P(S_i \le (2p(1-t)-1)N_1) &\le e^{-\frac{t^2p^2N_1}{2}},\text{ for all }0 < t < 1.
\end{split}
\end{eqnarray}
Taking $t = q / c$ with $q := \sqrt 2 / p$ for an appriprate positive constant $c$ to be specified later, we obtain for any $i$, it holds w.p $1-e^{-N_1/c}$ that
$$
S_i \ge ((2p-1)-q/c) N_1 = (a-q/c)N_1,
$$
where $a := 2p-1 = 1-\theta \in (1/2,1]$ as before.
A union bound then over $i \in [k]$ then gives: w.p $1-\delta(N_1) = 1 - ke^{-N_1/c}$ it holds that
\begin{eqnarray}
\frac{Z(\mathcal T)}{(aN_1)^k} \ge \left(1-\frac{q/a}{c}\right)^k = \left(\left(1-\frac{q/a}{c}\right)^{c}\right)^{k/c} \ge e^{-(q/a)k/c}.
\end{eqnarray}

Now, we want $k$ to be as large as possible, and the RHS of the above to be as large as possible too. We can achieve this by ensuring that

- $\delta(N_1) = e^{-N_1/c + \log k} = e^{-N_1/c+ \log N - \log N_1} \to 0$, and

To satisfy the above constraint (perhaps non-optimally!) it suffices to take
\begin{eqnarray}
N_1 = C\log N,\,k=N/N_1 = N/(C\log N),
\end{eqnarray}
with $C > c$ to be tuned later. Then, $\delta(N_1) = e^{-(C/c) \log N + \log N - \log\log N - \log C} = 1/N^{C/c-1}=o(1)$, $k/c = N/(Cc\log N)$ and it holds w.p $1-o(1)$ that
$$
Z(\mathcal T)/(aN_1)^k \gtrsim  b^{N/(C\log N)}
$$
where $b := e^{-(q/(ac))} \in (0,1)$. Rearranging, we deduce that
\begin{eqnarray}
    \begin{split}
Z(\mathcal T) &\gtrsim (a N_1)^kb^{N/\log N} = (aC \log N)^{N/(C\log N)}b^{N/\log N}\\
&= (aCb^{1/\log N} \log N)^{N/(C\log N)} \gtrsim ((aC/2) \log N)^{N/(C\log N)},
    \end{split}
\end{eqnarray}
w.p $1-o(1)$, which proves the claim. $\quad\quad \Box$