On a circumference, initially, 1 and 2 are written at opposite positions. Each operation involves writing the sum of adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. After performing the operation sufficiently many times, find the sum of the number of 2015s and 2016s written.
Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking for should be
$$
\varphi(2015)+\varphi(2016)=2016.
$$

Since the number written each time is the sum of the two numbers on its sides, observing the numbers on the circle in a clockwise direction, it is only necessary to prove that for any positive integer $n$ and any positive integer $k$ coprime with $n$, $n$ is written as the sum of its left neighbor $k$ and right neighbor $n-k$ exactly once.
We will prove this using the second principle of mathematical induction.
When $n=1,2,3$, it is obviously true.
When $n>3$, since $k$ and $n-k$ are coprime, they are not equal.

If $k>n-k$, according to the induction hypothesis, $k$ is written exactly once as the sum of its left neighbor $2k-n$ and right neighbor $n-k$;

If $k<n-k$, according to the induction hypothesis, $n-k$ is written exactly once as the sum of its left neighbor $k$ and right neighbor $n-2k$.
Therefore, by the second principle of mathematical induction, the conclusion holds.