Let's suppose $a\_i \sim \mathcal{U}([-N,N])$ where $[-N,N] \subset \mathbb{Z}$. We may then define: 


\begin{equation}
S\_n = \sum\_{i=1}^n a\_i \tag{1}
\end{equation}


Now, in order to estimate $\lvert H\_{2n} \rvert$ we may try to find an asymptotic estimate of: 


\begin{equation}
P(S\_{2n}=0) \tag{2}
\end{equation}


By decomposing $S\_n$ into positive and negative parts: 


\begin{equation}
S\_n = S\_n^+ + S\_n^- \tag{3}
\end{equation}


where $S\_n^+$ defines the sum of positive terms and $S\_n^-$ defines the sum of negative terms I reasoned that the average positive and negative step length should be approximately $\Delta = \frac{N}{2}$ when $n$ is large so:


\begin{equation}
P(S\_{2n}=0) \sim \frac{1}{2^{2n}} {2n \choose n} \sim \frac{1}{2^{2n}} \frac{\sqrt{4 \pi n}(\frac{2n}{e})^{2n}}{2 \pi n (\frac{n}{e})^{2n}} \sim \frac{1}{\sqrt{n}} \tag{4}
\end{equation}


This would imply that: 


\begin{equation}
\lvert H\_{2n} \rvert \sim \frac{(2N+1)^{2n}}{\sqrt{n}} \tag{5}
\end{equation}


but I must admit that my reasoning wasn't very rigorous here. Might there be a rigorous estimate of $\lvert H\_{2n} \rvert$ using a probabilistic method? 


**Note:** $N$ is assumed to be fixed in the asymptotic regime. 



Asymptotic estimate of the cardinality of $H_{2n}=\{\vec{a} \in [-N,N]^{2n}:\sum_{i=1}^{2n} a_i = 0\}$
According to the local central limit theorem (see e.g. [Esseen, Theorem 5, page 63](https://projecteuclid.org/euclid.acta/1485888404)), for any fixed natural $N$,
$$|H\_{2n}|\sim\frac{(2N+1)^{2n}}{2s\sqrt{\pi n}}
$$
as $n\to\infty$, where $s$ is the standard deviation of the uniform distribution on the set $\{-N,\dots,N\}$. 


(The definition of the condition $(L\_d)$ used in the mentioned theorem by Esseen is given at the bottom of page 54 of Esseen's paper.)


