Find $n$ such that $n\sqrt5 - \lfloor{n\sqrt5}\rfloor$ is maximised or minimised?
Locating a real number in the Stern–Brocot tree gives good rational approximations with increasing denominators.
For $\sqrt5$, below is the output for denominators at most $2009$. The last line says that the best approximations with this restriction on denominators are $3571/1597$ and $2889/1292$. The denominators in these two fractions are the ones you seek. You just need to test which is which.
$$ \begin{array}{rrrrr} n& a& b& c& d& \\ 1& 1& 1& 1& 0 \\ 2& 2& 1& 1& 0 \\ 3& 2& 1& 3& 1 \\ 4& 2& 1& 5& 2 \\ 5& 2& 1& 7& 3 \\ 6& 2& 1& 9& 4 \\ 7& 11& 5& 9& 4 \\ 8& 20& 9& 9& 4 \\ 9& 29& 13& 9& 4 \\ 10& 38& 17& 9& 4 \\ 11& 38& 17& 47& 21 \\ 12& 38& 17& 85& 38 \\ 13& 38& 17& 123& 55 \\ 14& 38& 17& 161& 72 \\ 15& 199& 89& 161& 72 \\ 16& 360& 161& 161& 72 \\ 17& 521& 233& 161& 72 \\ 18& 682& 305& 161& 72 \\ 19& 682& 305& 843& 377 \\ 20& 682& 305& 1525& 682 \\ 21& 682& 305& 2207& 987 \\ 22& 682& 305& 2889& 1292 \\ 23& 3571& 1597& 2889& 1292 \\ \end{array} $$ Here is Python code to generate this table:
from math import sqrt
t=sqrt(5)
a,b=0,1
c,d=1,0
n=0
while 1:
n=n+1
e=a+c
f=b+d
s=(e+0.0)/f
if s<t:
a,b=e,f
else:
c,d=e,f
print(n,a,b,c,d)
if b>2009 or d>2009:
break
Another method is to use the familiar Fibonacci approximants for $\phi=(1+\sqrt{5})/2$. Rendering $\sqrt{5}=2\phi-1$, carry the upper bounds until you get a maximum odd denominator $\le 2009$, or a maximum even denominator $\le 2×2009$, and take whichever is later:
$\frac{2}{1},\frac{5}{3},\frac{13}{8},...\frac{1597}{987},\color{blue}{\frac{4181}{2584}}$
Do the same with the lower bounds:
$\frac{1}{1},\frac{3}{2},\frac{8}{5},...\frac{987}{610},\color{blue}{\frac{2584}{1597}}$
So the optimal bounds within the problem constraints are:
$\frac{2584}{1597}<\phi<\frac{4181}{2584}$
and with $\sqrt{5}=2\phi-1$:
$\frac{3571}{1597}<\sqrt{5}<\frac{2889}{1292}$.