smc

Since, $x+y=26$, $x$ can equal $15$, and $y$ can equal $11$, so no even integers are required to make 26. To get to $41$, we have to add $41-26=15$. If $a+b=15$, at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$. Finally, we have the last transition is $57-41=16$. If $m+n=16$, $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\boxed{1}$. ~Extremelysupercooldude (Latex, grammar, and solution edits)