This is easy to determine by the known closed formulæ for $\dim \mathcal{S}_k\left(\Gamma_0(N)\right)$ and $\dim \mathcal{M}_k\left(\Gamma_0(N)\right)$ of Cohen and Oesterlé, namely for $k \in 1/2 + \mathbb{Z}$ with $k > 3/2$ and $N \in \mathbb{N}$, \begin{align} \dim \mathcal{S}_k\left(\Gamma_0(4N)\right) & = \frac{k - 1}{12} 4N \prod_{p|4N}\left(1 + \frac{1}{p}\right) - \frac{\zeta(k,4N)}{2} \prod_{\substack{p|4N \\ p > 2}} \lambda(r_p,s_p,p),\\ \dim \mathcal{M}_k\left(\Gamma_0(4N)\right) & = \frac{k - 1}{12} 4N \prod_{p|4N}\left(1 + \frac{1}{p}\right) + \frac{\zeta(2 - k,4N)}{2} \prod_{\substack{p|4N \\ p > 2}} \lambda(r_p,p), \end{align} where if $r_p$ is the power of $p$ dividing $4N$, then \begin{align} \lambda(r_p,p) &= \begin{cases} p^{r_p/2}\left(1 + \frac{1}{p}\right) & \text{if $r_p$ is even,}\\ 2 p^{(r_p - 1)/2} & \text{if $r_p$ is odd,} \end{cases}\\ \zeta(k,4N) & = \begin{cases} 2^{(r_2 + 1)/2} & \text{if $r_2 \geq 5$ is odd,}\\ 2^{r_2/2 + 1} & \text{if $r_2 \geq 4$ is even,}\\ 3 & \text{if $r_2 = 3$,}\\ 2 & \text{if $r_2 = 2$ and $r_p$ is odd for some prime $p|4N$ with $p \equiv 3\pmod{4}$,}\\ 3/2 & \text{if $r_2 = 2$, $r_p$ is even for all $p \equiv 3\pmod{4}$, and $k \in 1/2 + 2\mathbb{Z}$,}\\ 5/2 & \text{if $r_2 = 2$, $r_p$ is even for all $p \equiv 3\pmod{4}$, and $k \in 3/2 + 2\mathbb{Z}$.} \end{cases} \end{align} This is Théorème 2 of the paper of Cohen and Oesterlé, which reappears as Theorem 1.56 of The Web of Modularity: Arithmetic of the Coefficients of Modular Forms and Q-Series by Ken Ono (we take $\chi = \chi_0$ in this case, so that $s_p = 0$ for all $p$). As an aside, it must be pointed out that the proof of this theorem has actually never appeared in print!
From here it is clear that for $k \in \mathbb{N}$, $$\dim \mathcal{S}_{2k+1/2}\left(\Gamma_0(4N)\right) + \dim \mathcal{M}_{2k + 3/2}\left(\Gamma_0(4N)\right) = \frac{4kN}{3} \prod_{p|4N}\left(1 + \frac{1}{p}\right)$$ and similarly $$\dim \mathcal{S}_{2k+3/2}\left(\Gamma_0(4N)\right) + \dim \mathcal{M}_{2k + 1/2}\left(\Gamma_0(4N)\right) = \frac{4kN}{3} \prod_{p|4N}\left(1 + \frac{1}{p}\right),$$ as required.