Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$. 
Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.

This problem seems extremely hard until you find out that $f(x)+f(1-x)=1$ and you can then evalute $f(\frac{1}{1996})+f(\frac{1995}{1996}) =1, \dots, f(\frac{998}{1996})+f(1-\frac{998}{1996})=f(\frac12)+f(\frac12)=1 \implies f(\frac{998}{1996})=\frac12$. Thus the sum is just $997+\frac12$. My question is that how can one see something like this from a question like this without having gone through some contest math training where these kinds of "tricks" are taught. Is there some kind of tell here from which one can find this "property"?

There are two aspects which can lead us from
\begin{align*}
f(x)=\frac{9^x}{9^x+3}\tag{1}
\end{align*}
to the functional equation $f(x)+f(1-x)=1$.

Simplification: We can reduce the number of occurrences of $x$ in (1) by factoring out $9^x=3^{2x}$. We obtain
\begin{align*}
f(x)=\frac{9^x}{9^x+3}=\frac{1}{1+3^{1-2x}}\tag{2}
\end{align*}
Known pattern: The right hand side of (2) is a well known pattern which admits the following relationship
\begin{align*}
\frac{1}{1+a}+\frac{1}{1+\frac{1}{a}}=\frac{1}{1+a}+\frac{a}{1+a}=1\tag{3}
\end{align*}

The convenient relationship (3) applies to (2) with a nice twist of
\begin{align*}
\frac{1}{a}=\frac{1}{3^{1-2x}}=3^{-(1-2x)}=3^{1-2(1-x)}
\end{align*}
from which
\begin{align*}
\color{blue}{f(x)+f(1-x)=1}
\end{align*}
follows.
