Note that the $x$ and $y$ coordinates of these three points are reciprocals. Therefore, let's consider the function $g(x) = xf(x)-1$. Clearly, $g(x)$ is a $5^{th}$ degree polynomial equation. We have \begin{align} g(2)&=2f(2)-1=2\times\frac{1}{2}-1 = 0\\ g(3)&=3f(3)-1=3\times\frac{1}{3}-1 = 0\\ g(4)&=4f(4)-1=4\times\frac{1}{4}-1 = 0 \end{align} This means $2, 3, 4$ are the three roots of $g(x)$. Hence, $g(x)$ can be written as $$g(x)=(x-2)(x-3)(x-4)(x^2 +px +q)$$ where $p$ and $q$ are two coefficients. Setting $x=0$ yields $$g(0)= 0f(0) -1 = -1\ \text{and}\ (0-2)(0-3)(0-4)(0+0+q)=-1 \implies q=\frac{1}{24}$$ Now, setting $x=1$ and $x=5$, respectively, $$ \begin{array}{} g(1) &= f(1) - 1 &= -6\times (1+p+q)\\ g(5) &= 5f(5) - 1 &= 6\times (25+5p + q) \end{array} $$ Therefore, \begin{align} f(1) + f(5)&=(-6\cdot(1+p+q) + 1) +\big(\frac{1}{5}\cdot 6\cdot(25+5p+q)+1\big)\\ &=25 + \frac{1}{5} -\frac{24}{5}q\\ & = 25 + \frac{1}{5} -\frac{24}{5}\times\frac{1}{24}\\ &=\boxed{25} \end{align}