First, rewrite it a bit: $rem \to mod$. $\left\lfloor\frac{n-i}{i}\right\rfloor=\left\lfloor\frac{n}{i}-1\right\rfloor=\left\lfloor\frac{n}{i}\right\rfloor-1$
$$T(n,i) = \left(\left\lfloor\frac{n}{i}\right\rfloor-1\right)i + T\left(i +(n\operatorname{mod}i), (n \operatorname{mod} i)\right)$$
Consider $n=ki+l$, where $k-$non-negative integer, $l-$integer such that $0\le l
Then,
$$
T(n,i)= (k-1)i+T(i+l,l)=n-l-i+T(i+l,l)
$$
Now, consider $i=ml+h$, where $m-$non-negative integer, $h-$integer, $0\le h
Then,
$$
T(i+l,l)=ml+T(l+h,h)=i-h+T(l+h,h)
$$
So, we've got repetitive pattern, which ends when we encouter $T(x,1)$ or $T(x,0)$.
Let us denote $i=a_0$, $l=a_1$, $h=a_2$ and so on, where general term is $a_{p+1}=a_{p-1}\operatorname{mod}a_p$, then
$$
\begin{align*} T(n,i) &= n-a_0-a_1+T(a_0+a_1,a_1)=\\
&= n-a_1-a_2+T(a_1+a_2,a_2)=\\
&= n-a_2-a_3+T(a_2+a_3,a_3)=\\
&= \dots =\\
&= n-a_{p-1}-a_p+T(a_{p-1}+a_p,a_p)
\end{align*}
$$
Hence, in case of $\color{red}{a_p=1}\to T(a_{p-1}+a_p,a_p)=a_{p-1}+a_p$
$$
\color{red}{T(n,i)=n}
$$
In case of $\color{red}{a_p=0}\to T(a_{p-1}+a_p,a_p)=0$
$$
\color{red}{T(n,i)=n-a_{p-1}\to O(n)}
$$
Notice, that in last case $2\le a_{p-1}
Thus, we can combine both cases
$$
\color{red}{T(n,i)\to O(n),\quad C\in[1/2,1]}
$$
where $C$ is a constant hidden behind $O(n)$.