(i) To solve the differential equation \(\frac{dN}{dt} = \frac{N(1800 - N)}{3600}\), we first separate variables:

\(\frac{1}{N(1800 - N)} = \frac{1}{3600} \frac{dN}{dt}.\)

We use partial fraction decomposition:

\(\frac{1}{N(1800 - N)} = \frac{A}{N} + \frac{B}{1800 - N}.\)

Solving for \(A\) and \(B\), we find:

\(\frac{2}{N} + \frac{2}{1800 - N}.\)

Integrating both sides gives:

\(2 \ln N - 2 \ln (1800 - N) = \frac{t}{2} + C.\)

Using the initial condition \(N = 300\) when \(t = 0\), we solve for \(C\):

\(2 \ln 300 - 2 \ln 1500 = C.\)

Thus, the solution is:

\(N = \frac{1800e^{t/2}}{5 + e^{t/2}}.\)

(ii) As \(t \to \infty\), \(e^{t/2} \to \infty\), so \(N \to 1800\).