Step 1 (Model the profit per subscriber).
Let $n$ be the number of subscribers. For every extra subscriber above $725$, the profit per subscriber drops by ₹ $0.01$ \emph{for all subscribers}. Hence
\[
p(n)=12-0.01\,(n-725)=12-0.01n+7.25=\boxed{\,19.25-0.01n\,}\;(\text{rupees per subscriber}).
\]
Step 2 (Total profit as a function of $n$).
\[
P(n)=n\,p(n)=n(19.25-0.01n)=\boxed{-0.01n^2+19.25\,n},
\]
which is a concave quadratic (opens downward), so it has a unique maximum at its vertex.
Step 3 (Continuous maximizer / vertex).
For $P(n)=-an^2+bn$, the vertex is at $n^\star=\dfrac{b}{2a}$. Here $a=0.01,\ b=19.25$:
\[
n^\star=\frac{19.25}{2\cdot 0.01}=\frac{19.25}{0.02}=962.5.
\]
Since $n$ must be an integer, only the nearest integers $962$ and $963$ can maximize $P$.
Step 4 (Show the tie rigorously).
A discrete check via first difference:
\[
P(n{+}1)-P(n)=\left[-0.01(n{+}1)^2+19.25(n{+}1)\right]-\left[-0.01n^2+19.25n\right]
=19.24-0.02n.
\]
At $n=962$, $P(963)-P(962)=19.24-0.02\cdot 962=0 \Rightarrow P(963)=P(962)$ (tie).
Also $P(962)-P(961)=19.24-0.02\cdot 961 > 0$, so $P(962) > P(961)$.
Step 5 (Numerical confirmation).
\[
\begin{aligned}
P(962)&=962\big(19.25-9.62\big)=962\times 9.63=₹ 9264.06,
P(963)&=963\big(19.25-9.63\big)=963\times 9.62=₹ 9264.06.
\end{aligned}
\]
Both equal and exceed the profit at neighboring integers.
\[
\boxed{\text{Maximum total profit occurs at } n=962 \text{ \emph{and} } n=963.}
\]