Example 1 (32nd Russian Mathematical Olympiad) The product $a_{1} \cdot a_{2} \cdot \cdots \cdot a_{100}$ is written on the blackboard, where $a_{1}, a_{2}, \cdots, a_{100}$ are positive integers. If one of the multiplication signs is changed to an addition sign (keeping the other multiplication signs), it is found that 32 of the 99 resulting sums are even. How many even numbers can there be at most among $a_{1}, a_{2}, \cdots, a_{100}$?
Solve 33.

Let the leftmost even number in $a_{1}, a_{2}, \cdots, a_{100}$ be $a_{i}$, and the rightmost even number be $a_{k}$. Denote $X_{j}=a_{1} a_{2} \cdots a_{j}, Y_{j}=a_{j+1} a_{j+2} \cdots a_{100}$.
It is easy to see that when $j=1,2, \cdots, i-1$, $X_{j}$ is odd and $Y_{j}$ is even, so the sum $X_{j}+Y_{j}$ is odd; when $j=k, k+1, \cdots, 100$, $X_{j}$ is even and $Y_{j}$ is odd, and the sum $X_{j}+Y_{j}$ is also odd.

Only when $j=i, i+1, \cdots, k-1$, both $X_{j}$ and $Y_{j}$ are even, and the sum $X_{j}+Y_{j}$ is even. This indicates that $k-i=32$.

Since the numbers between $a_{i}$ and $a_{k}$ can be either odd or even, the number of even numbers in $a_{1}, a_{2}, \cdots, a_{100}$ is maximized only when they are all even. Therefore, there can be at most 33 even numbers.