An easy one. Just write $n=a^2+b,\ b\in [0;2a+1)$ and we have that $a-2|a^2+b-4$ and $a+2|a^2+b+4$ must hold simultaneously. However estimating the quotient from the first equation we get one of the following possibilities: $b=0$, $b=a-2$ or $b=2a-4$ (maybe some extra checking of small values of $a$ is needed). However having these we can handle the problem easily analysing the second divisibility relation.

Let $ n=a^2+b$, $ a \in \mathbb{N}$, $ b \in \left(\left[1,\ 2a\right] \cap \mathbb{N}\right) \cup \left\{0\right\}$. $ \mathbb{N}$ represents the set of positive integers. $ \Longrightarrow$ \[ \underbrace{\left\lfloor\sqrt{n}\right\rfloor-2 \mid n-4 \Longleftrightarrow a-2 \mid a^2 + b - 4}_{\left(1\right)} \quad \underbrace{\left\lfloor\sqrt{n}\right\rfloor+2 \mid n+4 \Longleftrightarrow a+2 \mid a^2 + b+ 4}_{\left(2\right)}\] Now, from $ \left(1\right)$ it's obvious that $ a \neq 2$, for $ a=1$, from $ \left(2\right)$ we have $ b=1$. Thus, one solution is $ n=2$. $ a=3 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 6\right\},\ \left(2\right) \Longrightarrow 5 \mid 13 + b \Longrightarrow b = 2 \Longrightarrow n=11$. $ a=4 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 8\right\},\ \left(1\right) \Longrightarrow 2 \mid 12 + b \Longrightarrow 2 \mid b,\ \left(2\right) \Longrightarrow 6 \mid 20 + b$ $ \Longrightarrow b=4 \Longrightarrow n=20$. $ a=5 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 10\right\},\ \left(1\right) \Longrightarrow 3 \mid 21 + b \Longrightarrow 3 \mid b,\ \left(2\right) \Longrightarrow 7 \mid 29 + b$ $ \Longrightarrow b=6 \Longrightarrow n=31$. $ a=6 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 12\right\},\ \left(1\right) \Longrightarrow 4 \mid 32 + b \Longrightarrow 4 \mid b,\ \left(2\right) \Longrightarrow 8 \mid 40 + b$ $ \Longrightarrow b=0,\, 8 \Longrightarrow n=36,\, 44$. For $ a>6$ we have $ \forall m \in \mathbb{N} \setminus \left\{1,\, 2\right\}$, $ m\left(a-2\right)>2a$. Now, $ \left(1\right) \Longrightarrow a-2 \mid \left(a-2\right)\left(a+2\right)+b$, thus $ b \in \left\{0,\, a-2,\, 2\left(a-2\right)\right\}$. $ b=0,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 4 \Longrightarrow a + 2 \mid \left(a+2\right)^2 -4\left(a+2\right)+8 \Longrightarrow a+2 \mid 8$, but $ a+2 > 8$, no solution! $ b=a-2,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + a + 2 \Longrightarrow a+2 \mid \left(a+2\right)^2-4\left(a+2\right)+4 \Longrightarrow a+2 \mid 4$, no solution! $ b=2a-4,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 2a \Longrightarrow \forall a>6,\ n=a^2+2a-4$ is solution. Finally, solutions are $ n=2,\, 36,\, a^2+2a-4$, where $ a \in \mathbb{N} \setminus \left\{1,\, 2\right\}$.

$ n = a^{2} + k$ ,$ k\leq 2a$ $ a - 2|a^{2} + k - 4$ ,$ \implies$ $ a - 2|k$ ,$ a + 2|a^{2} + k + 4$ $ \implies$ $ a + 2|k + 8$ $ a > 6$ $ \implies$ $ k = 0,k = a - 2,k = 2a - 4$ $ i)k = 0, a + 2|8$ $ \implies$ but $ a > 6$ contradiction! $ ii)k = a - 2$ $ \implies$ $ n = a^{2} + a - 2, a + 2|a^{2} + a + 2$ $ a + 2|a^{2}$ $ \implies$ $ a + 2|4$ but $ a > 6$ contradiction! $ iii)k = 2a - 4$ $ \implies$ $ a + 2|a^{2} + 2a$ $ n=a^{2}+2a-4$ $ (a>6)$ $ a\leq 6$ $ \implies$ ,$ n = 2,n = 36$

JANKRI wrote: Let $ n=a^2+b$, $ a \in \mathbb{N}$, $ b \in \left(\left[1,\ 2a\right] \cap \mathbb{N}\right) \cup \left\{0\right\}$. $ \mathbb{N}$ represents the set of positive integers. $ \Longrightarrow$ \[ \underbrace{\left\lfloor\sqrt{n}\right\rfloor-2 \mid n-4 \Longleftrightarrow a-2 \mid a^2 + b - 4}_{\left(1\right)} \quad \underbrace{\left\lfloor\sqrt{n}\right\rfloor+2 \mid n+4 \Longleftrightarrow a+2 \mid a^2 + b+ 4}_{\left(2\right)}\] Now, from $ \left(1\right)$ it's obvious that $ a \neq 2$, for $ a=1$, from $ \left(2\right)$ we have $ b=1$. Thus, one solution is $ n=2$. $ a=3 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 6\right\},\ \left(2\right) \Longrightarrow 5 \mid 13 + b \Longrightarrow b = 2 \Longrightarrow n=11$. $ a=4 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 8\right\},\ \left(1\right) \Longrightarrow 2 \mid 12 + b \Longrightarrow 2 \mid b,\ \left(2\right) \Longrightarrow 6 \mid 20 + b$ $ \Longrightarrow b=4 \Longrightarrow n=20$. $ a=5 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 10\right\},\ \left(1\right) \Longrightarrow 3 \mid 21 + b \Longrightarrow 3 \mid b,\ \left(2\right) \Longrightarrow 7 \mid 29 + b$ $ \Longrightarrow b=6 \Longrightarrow n=31$. $ a=6 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 12\right\},\ \left(1\right) \Longrightarrow 4 \mid 32 + b \Longrightarrow 4 \mid b,\ \left(2\right) \Longrightarrow 8 \mid 40 + b$ $ \Longrightarrow b=0,\, 8 \Longrightarrow n=36,\, 44$. For $ a>6$ we have $ \forall m \in \mathbb{N} \setminus \left\{1,\, 2\right\}$, $ m\left(a-2\right)>2a$. Now, $ \left(1\right) \Longrightarrow a-2 \mid \left(a-2\right)\left(a+2\right)+b$, thus $ b \in \left\{0,\, a-2,\, 2\left(a-2\right)\right\}$. $ b=0,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 4 \Longrightarrow a + 2 \mid \left(a+2\right)^2 -4\left(a+2\right)+8 \Longrightarrow a+2 \mid 8$, but $ a+2 > 8$, no solution! $ b=a-2,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + a + 2 \Longrightarrow a+2 \mid \left(a+2\right)^2-4\left(a+2\right)+4 \Longrightarrow a+2 \mid 4$, no solution! $ b=2a-4,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 2a \Longrightarrow \forall a>6,\ n=a^2+2a-4$ is solution. Finally, solutions are $ n=2,\, 36,\, a^2+2a-4$, where $ a \in \mathbb{N} \setminus \left\{1,\, 2\right\}$. It's possible to have $n=4$ here. In fact, $$\left\lfloor\sqrt{4}\right\rfloor-2 \mid 4-4 $$(because $0|0$) and $$\left\lfloor\sqrt{4}\right\rfloor+2 \mid 4+4 .$$Moreover, we can say that we can have $a=2$.