First, we'll need a few lemmas:
Lemma 1: If $a$, $b$ are coprime, then
$$\mathrm{gnu}(ab) \geq \mathrm{gnu}(a) \mathrm{gnu}(b).$$
Sketch of proof: Let $G_1, \dots G_n$ be nonisomorphic groups of order $a$, $H_1, \dots, H_m$, be nonisomoprhic groups of order $b$. Then we can show $(G_i \times H_j)_{i=1,...,n}^{j=1,...,m}$ are all nonisomorphic due to the coprimality of $a$ and $b$.
In fact, it is conjectured that the above holds for all $a, b$.
Lemma 2: For $p$ prime, $q$ coprime to $p$, $m > 1$, $\mathrm{gnu}(p^mq) \geq \mathrm{gnu}(p^{m-1}q)$. In particular, $\mathrm{gnu}(2^m) \geq 5$ for all $m \geq 3$.
Sketch of proof: Can be proved inductively by showing that the groups of the previous prime power order, direct product the cyclic group of order $p$, are all non-isomorphic. Second statement follows immediately after observing $\mathrm{gnu}(8)=5$.
Lemma 3: For $p$ an odd prime, $\mathrm{gnu}(4p) \geq 4$.
Sketch of proof: One can show $C_{4p}$, $C_{2} \times C_2 \times C_p$, $D_{4p}$, $\mathrm{Dic}_{4p}$ are all non-isomorphic, where $C_m$ is the cyclic group of order $m$, $D_m$ is the dihedral group of order $m$, $\mathrm{Dic}_m$ is the dicyclic group of order $m$.
Lemma 4: For $p$ an odd prime, $t \geq 1$, $\mathrm{gnu}(4p^t) \geq 4$.
Proof: If $t = 1$, this is just lemma 3. If $t \geq 1$,
$$
\begin{align}
\mathrm{gnu}(4p^t) &\geq \mathrm{gnu}(4) \mathrm{gnu}(p^t) && \text{lemma 1}\\
&\geq \mathrm{gnu}(4) \mathrm{gnu}(p^2) && \text{lemma 2}\\
&\geq 2 \cdot 2 \\
&\geq 4.
\end{align}
$$
Now, we will successively show that $2 \mid n$, then $4 \mid n$, then $8 \mid n$, then $3 \mid n$.
$2 \mid n$:
In fact, this shows that if
$$\mathrm{gnu}(n+i) = i \quad i=1,...,2$$
holds, then $2 \mid n$.
Since $\mathrm{gnu}(n+1) = 1$, we have that either $n+1=2$ or $n+1$ is odd, as there exists a dihedral group of order $2m$ for all $m>2$, and $\mathrm{gnu}(4) = 2$.
But if $n=1$, $\mathrm{gnu}(n+2) = \mathrm{gnu}(3) = 1$.
So $n$ is even.
$4 \mid n$:
In fact, this shows that if
$$\mathrm{gnu}(n+i) = i \quad i=1,...,3$$
holds, then $4 \mid n$.
We already have $2 \mid n$, so suppose for contradiction $n \equiv 2 \mod 4$. Then $n + 2 \equiv 0 \mod 4$.
Write $n = 2^rs$ where $s$ is odd. We have $r \geq 2$. If $r \geq 3$
$$\begin{align}
\mathrm{gnu}(n+2) &= \mathrm{gnu}(2^r s) \\
&\geq \mathrm{gnu}(2^r)\mathrm{gnu}(s) && \text{lemma 1} \\
&\geq \mathrm{gnu}(2^r) \\
&\geq 5 && \text{lemma 2} \\
\end{align}$$
which is a contradiction. $(\ast)$
So $r = 2$. As $m$ is odd, either $m=1$ or there is some odd prime $p \mid m$. Let $p^t$ be the largest power of $p$ dividing $m$, with $t \geq 1$. Then $n+2 = 4p^t u$ with $u$ coprime to $4, p^t$ and hence
$$\begin{align}
\mathrm{gnu}(n+2) &= \mathrm{gnu}(4p^t u) \\
&\geq \mathrm{gnu}(4p^t) \mathrm{gnu}(u) && \text{lemma 1} \\
&\geq \mathrm{gnu}(4p^t) \\
&\geq 4 && \text{lemma 4}
\end{align}$$
which is also a contradiction.
So, $4 \mid n$.
$8 \mid n$:
We have $4 \mid n$. Suppose $n = 4m$, for $m$ odd. Then $8 \mid n+4$. It follows similarly to how we previously argued at $(\ast)$ that $\mathrm{gnu}(n+4) \geq 5$, a contradiction. Hence $8 \mid n$.
$3 \mid n$:
We have $n = 8m$. Suppose $3 \not \mid m$ for contradiction.
Then $n+4 = 4(2m+1)$. $2m+1 \not \equiv 1 \mod 3$, as $m \not \equiv 0 \mod 3$.
Hence we can rewrite $n+4 = 4r$, $r$ odd, $r \not \equiv 1 \mod 3$.
If $r \equiv 0 \mod 3$, $12 \mid n+4$. Similarly to how we have previously argued using lemma 1 and lemma 2, we can show $\mathrm{gnu}(n+4) \geq \mathrm{gnu}(12) = 5$, a contradiction.
If $r \equiv 2 \mod 3$, then $n+4 = 4(3s+2) = 12s+8$ for some $s$.
But $n = 12s+4$, and $8 \mid n$, so we require $s = 2t+1$ is odd.
Hence $n = 12(2t+1)+4 = 24t+16$.
So $n+2 = 24t+18 = 6(4t+3)$.
If $3 \mid t$, then $18 \mid n+2$. Again we can argue with lemma 1 and lemma 2 that $\mathrm{gnu}(n+2) \geq \mathrm{gnu}(18) = 5$, a contradiction.
Otherwise, we have $n+2 = 6u$, for some $u$ with $2, 3 \not \mid u$. So there is a prime $p$ such that $6p \mid n+2$. Yet again, we argue with lemma 1 and 2 that $\mathrm{gnu}(n+2) \geq \mathrm{gnu}(6p)$.
But then, $C_2 \times C_3 \times C_p$, $C_3 \times D_{2p}$, $C_p \times D_6$, $D_{6p}$ can all be shown to be non-isomorphic groups of order $6p$. Thus, $\mathrm{gnu}(n+2) \geq 4$, another contradiction.
Hence, we have $3 \mid n$.
As $3 \mid n$ and $8 \mid n$, we have $24 \mid n$ as required.
