Find the positive integer $n$ such that 
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]
Since we are dealing with acute angles, $\tan(\arctan{a}) = a$.
Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$, by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$.
Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$.
Now, $\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}$.
We now have $\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}$. Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1$; and simplifying, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}$.