$7 \cdot 82$ Given 155 birds are perched on a circle $C$. If $\overparen{P_{i} P_{j}} \leqslant 10^{\circ}$, then the birds $P_{i}$ and $P_{j}$ are said to be mutually visible. Find the minimum number of pairs of mutually visible birds (you can assume that multiple birds can occupy the same position).
[Solution] Let birds $\alpha$ and $\beta$ be located at points $A$ and $B$ on circle $C$, respectively, and assume that birds $\alpha$ and $\beta$ can see each other. Let the number of birds that can be seen from point $B$ but not from point $A$ be $k$, and the number of birds that can be seen from point $A$ but not from point $B$ be $h$. Without loss of generality, assume $k \geqslant h$. It is easy to see that if all the birds at point $B$ are moved to point $A$, the number of visible pairs of birds will not increase. However, this reduces the number of points with birds by 1. Repeating this process until all visible birds are at the same point, the number of points with birds will be at most 35. If not, there would be at least 36 points with birds, and among these points, there must be two points with an arc length of no more than $10^{\circ}$, making the birds at these two points visible to each other, which is a contradiction. On the other hand, let $A_{1}, A_{2}, \cdots, A_{35}$ be the vertices of a regular 35-sided polygon inscribed in circle $C$. Then, the birds at $A_{i}$ and $A_{j} (i \neq j)$ are not visible to each other. If the number of bird points $n \leqslant 34$, then the $n$ points can be moved to $A_{1}, A_{2}, \cdots, A_{n}$. Since $n \leqslant 34$, point $A_{35}$ will have no birds. If $A_{i}$ has at least two birds, then moving one bird from $A_{i}$ to $A_{35}$ will reduce the number of visible pairs. Continuing this process, we can ensure that there are exactly 35 points with birds, and these points are $A_{1}, A_{2}, \cdots, A_{35}$.

If there exist points $A_{i}$ and $A_{j} (i \neq j)$ with bird counts $x_{i}$ and $x_{j}$, respectively, and $x_{i} \geqslant x_{j} + 2$, then moving one bird from $A_{i}$ to $A_{j}$ will reduce the number of visible pairs by at least 1. Continuing this process, we can ensure that the difference between any two of $x_{1}, x_{2}, \cdots, x_{35}$ does not exceed 1. It is clear that when 20 of these 35 numbers are 4 and 15 are 5, the number of visible bird pairs reaches its minimum value $20 \binom{4}{2} + 15 \binom{5}{2} = 270$.