Solution:

a. Let $n \geq 4011^{2}$ and $m \in \mathbb{N}$ be such that $m^{2} \leq n < (m+1)^{2}$. Then
$$\begin{aligned} \left(1+\frac{1}{2005}\right) n - (m+1)^{2} & \geq \left(1+\frac{1}{2005}\right) m^{2} - (m+1)^{2} \\ & = \frac{m^{2}}{2005} - 2m - 1 \\ & = \frac{1}{2005}\left(m^{2} - 4010m - 2005\right) \\ & = \frac{1}{2005}\left((m-2005)^{2} - 2005^{2} - 2005\right) \\ & \geq \frac{1}{2005}\left((4011-2005)^{2} - 2005^{2} - 2005\right) \\ & = \frac{1}{2005}\left(2006^{2} - 2005^{2} - 2005\right) \\ & = \frac{1}{2005}(4011 - 2005) = \frac{2006}{2005} > 0 \end{aligned}$$
Thus we get
$$n < (m+1)^{2} < \left(1 + \frac{1}{2005}\right) n$$
and $l^{2} = (m+1)^{2}$ is the desired square.

b. We show that $M = 4010^{2} + 1$ is the required least number. Suppose $n \geq M$. Write $n = 4010^{2} + k$, where $k$ is a positive integer. Note that we may assume $n < 4011^{2}$ by part (a). Now
$$\begin{aligned} \left(1 + \frac{1}{2005}\right) n - 4011^{2} & = \left(1 + \frac{1}{2005}\right)\left(4010^{2} + k\right) - 4011^{2} \\ & = 4010^{2} + 2 \cdot 4010 + k + \frac{k}{2005} - 4011^{2} \\ & = (4010 + 1)^{2} + (k - 1) + \frac{k}{2005} - 4011^{2} \\ & = (k - 1) + \frac{k}{2005} > 0 \end{aligned}$$
Thus we obtain
$$4010^{2} < n < 4011^{2} < \left(1 + \frac{1}{2005}\right) n$$
We check that $M = 4010^{2}$ will not work. For suppose $n = 4010^{2}$. Then
$$\left(1 + \frac{1}{2005}\right) 4010^{2} = 4010^{2} + 2 \cdot 4010 = 4011^{2} - 1 < 4011^{2}$$
Thus there is no square integer between $n$ and $\left(1 + \frac{1}{2005}\right) n$.
This proves (b).