Let, $$S_n = \sum_{r=1}^{n}\sin^{-1}\left(\frac{1}{\sqrt{r^2+1}}\right)$$
If $n=1$ then, $S_1=\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^o$
If $n=3$ then, $S_3=\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+ \sin^{-1}\left(\frac{1}{\sqrt{10}}\right)+ = 90^o$
From two equations we get, $S_3-S_1=45^o$
We introduce the new variable $A_n=S_{\frac{n(n+1)}{2}} - S_{\frac{n(n-1)}{2}}$ when, $A=45^o, n=1 $ and $n=2$ satisfies the condition.
If $n=1 \implies S_{\frac{1(1+1)}{2}} - S_{\frac{1(1-1)}{2}} =S_1-S_0=45^o$
If $n=2 \implies S_{\frac{2(2+1)}{2}} - S_{\frac{2(2-1)}{2}} =S_3-S_1=45^o$
But $n=3$ does not satisfies the condition.
I want to find the sequence of natural numbers satisfies the condition, if $n \gt 2$ satisfies this condition.
There are no other such $n$.
It's easier to do the problem in radians.
If $\theta_r=\sin^{-1}\frac{1}{\sqrt{r^2+1}}$, then $\tan(\theta_r)=\frac{1}{r}.$ So:
$$S_n=\sum_{r=1}^{n}\tan^{-1}\frac{1}{r}.$$
Since $\tan \theta>\theta$ (in radians) when $\theta\in(0,\pi/2)$, we have $\tan^{-1}x<x$ for all $x>0$. If $T_n=\frac{n(n+1)}{2}$, we have
$$\begin{align}A_n=\sum_{r=T_{n-1}+1}^{T_n} \tan^{-1}\frac{1}{r}&<\sum_{r=T_{n-1}+1}^{T_n}\frac{1}{r}\\ &<\int_{T_{n-1}+1}^{T_n+1}\frac{dx}{x}\\&=\ln\frac{T_{n}+1}{T_{n-1}+1}\\ &=\log\left(1+\frac{n}{T_{n-1}+1}\right)\\ &<\frac{n}{T_{n-1}+1}\\&<\frac{2}{n-1} \end{align}$$
Now when $n\geq 4$, we get $$A_n<\frac{2}{n-1}\leq \frac{2}{3}<\frac{\pi}{4}.$$ since $3\pi>8.$
Since you've already checked $n=1,2,3$, we are done. The only such $n$ are $n=1,2.$
More generally, if $r_1,\dots,r_k$ are integers and $\theta=\sum_{j=1}^{k}\tan^{-1}\frac{1}{r_j}$ then $\tan(\theta)=1$ if and only if:
$$\prod_{j=1}^{k}(r_j+i) = m+mi$$ for some integer $m.$
For example, in your case $n=2$, we have $(2+i)(3+i)=5+5i.$
Another example is $r_1=2,r_2=4,r_3=13.$ Then $(2+i)(4+i)(13+i)=85+85i.$ So: $$\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{13}=45^\circ.$$ Other examples:
$$(2,5,8),\\(2,6,7,68),\\(2,4,15,98),\\(2,4,30,33,76),\\(2,6,8,44,105)\\(2,7,8,23,83),\\(3,3,11,32,50),\\(2,8,10,18,32,91),\\(2,8,12,18,37,46,100)$$
We can get infinitely many such solutions because we have that:
$$\tan^{-1}\dfrac{1}{n+\frac{n^2+1}{k}}=\tan^{-1}\frac1n-\tan^{-1}\frac1{n+k}$$
This means when $k\mid r_j^2+1$ we can replace any $r_j$ with the pair $r_j+k,r_j+\frac{r_j^2+1}{k}.$ to get a longer sequence.
More generally, if $k\mid r_i^2+1$ we can replace $r_i$ with $n+k$ and $n+\frac{r_i^2+1}{k}$. In particular, when $k=1$ we can always replace $r_j$ with $r_j+1,r_j^2+r_j+1.$
Indeed, it begins to look like all such examples can be built this way:
$$\begin{align}(1)&\to (2,3)\tag{r=1,k=1}\\ &\to (2,5,8)\tag{r=3,k=2}\\ &\to(2,7,8,18)\tag{r=5,k=2}\\ &\to(2,8,12,17,18)\tag{r=7,k=5}\\ &\to(2,8,12,18,27,46)\tag{r=17,k=10}\\ &\to (2,8,12,18,37,46,100)\tag{r=27,k=10}\end{align}$$
You can reduce backwards as follows. If $a,b$ are in your sequence, such that $a+b\mid ab-1$, then you can remove $a,b$ and replace with $\frac{ab-1}{a+b}.$ I'm not sure can always do this, however...