Asymptotics of of $T(n) = n + T(n - \sqrt{n})$
You can get a lower bound of $\Omega(n^{3/2})$ as follows. Consider the number of steps it takes to get from $n$ to $n/2$. Since at each step, $n$ decreases by at most $\sqrt{n}$, this takes at least $\sqrt{n}/2$ steps. At each such step, we accumulate at least $n/2$, for a total of at least $n^{3/2}/4$.
You can prove a matching upper bound using a similar approach. While the input parameter is at least $n/2$, it decreases by at least $\sqrt{n/2}$, and so it takes at most $\sqrt{n/2}$ steps to get to $n/2$. During each such step, we accumulate at most $n$, for a total of at most $n^{3/2}/\sqrt{2}$. An identical argument shows that at most $(n/2)^{3/2}/\sqrt{2}$ is accumulated while going from $n/2$ to $n/4$, and so on. Hence
$$
T(n) \leq \frac{n^{3/2}}{\sqrt{2}} \left(1 + \frac{1}{2^{3/2}} + \frac{1}{4^{3/2}} + \cdots\right) = O(n^{3/2}),
$$
since the geometric series inside the parentheses converges.
Numerical experiments suggest that
$$
T(n) \sim \frac{2}{3} n^{3/2}.
$$
We can show this heuristically as follows. Let us imagine that $T$ is a recurrence in continuous time. Then $T(n) - T(n-\sqrt{n}) \approx \sqrt{n} T'(n)$, and so
$$
T'(n) \approx \sqrt{n}.
$$
Integrating, we get
$$
T(n) \approx \frac{n^{3/2}}{3/2} = \frac{2}{3} n^{3/2}.
$$
This can probably be showed by running the argument above more carefully (replacing $n/2$ with $(1-\epsilon)n$).