After $4$ operations, the milk remaining is

$

80!\left(\frac{80-x}{80}\right)^{4}=\frac{(80-x)^{4}}{80^{3}}.

$

We need $25<\dfrac{(80-x)^{4}}{80^{3}}<30$, i.e., $25\cdot 80^{3}<(80-x)^{4}<30\cdot 80^{3}.$

$80^3 = 512 000$, so $1.28\times 10^{7}<(80-x)^{4}<1.536\times 10^{7}.$

Test integers $80-x$:

• $80-x=59:\ 59^{4}=12{,}117{,}361$ -- too small.
• $80-x=60:\ 60^{4}=12{,}960{,}000$ -- in range ($x=20$, milk $\approx 25.31$).
• $80-x=61:\ 61^{4}=13{,}845{,}841$ -- in range ($x=19$, milk $\approx 27.04$).
• $80-x=62:\ 62^{4}=14{,}776{,}336$ -- in range ($x=18$, milk $\approx 28.86$).
• $80-x=63:\ 63^{4}=15{,}752{,}961$ -- too large.

So $x\in\{18,19,20\}$. The largest such $x$ is $\boxed{20}$.

Why this is CAT-level: The hardening comes from the band condition combined with integer feasibility. The natural attempt -- setting up $(80-x)^4 = k\cdot 80^3$ for a clean $k$ -- fails because no clean closed-form root works; one must bracket the integer band by estimating $60^4$, $61^4$, $62^4$ mentally.

Answer: $x = 20$.