Example 25 (28th IMO Candidate Problem) Given $x=-2272, y=10^{3}+10^{2} c+10 b+a, z=1$ satisfy the equation $a x+b y+c z=1$, where $a, b, c$ are positive integers, $a<b<c$, find $y$.

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Given that $b(1000+100 c+10 b+a)+c-2272 a-1=0$.

If $b \geqslant a+2$, then $c \geqslant a+3$, from the above equation we get
$$
\begin{aligned}
0 & \geqslant(a+2)[100+100(a+3)+10(a+2)+a]+(a+3)-2272 a-1 \\
& =111 a^{2}-729 a+2642 .
\end{aligned}
$$

Let $u=111 a^{2}-729 a+2642$, i.e., $111 a^{2}-729 a+(2642-u)=0$, thus by $\Delta=729^{2}-4 \times$ $111 \times(2642-u) \geqslant 0$ we get
$$
u \geqslant \frac{4 \times 111 \times 2642-729^{2}}{4 \times 111}>0,
$$

which leads to a contradiction $0>0$, so $b=a+1$. Let $c=a+2+t$, then by $c>b=a+1$ we know $t \geqslant 0$, thus
$$
\begin{aligned}
0 & =(a+1)[1000+100(a+2+t)+10(a+1)+a]+a+2+t-2272 a-1 \\
& =111 a^{2}+(100 t-950) a+(1211-101 t) .
\end{aligned}
$$

If $t \geqslant 2$, then
$$
\begin{aligned}
0 & \geqslant 111 a^{2}+(200-950) a+(1211+202) \\
& =111 a^{2}-750 a+1413 .
\end{aligned}
$$

Let $v=111 a^{2}-750 a+1413$, i.e., $111 a^{2}-750 a+(1413-v)=0$, by $\Delta=750^{2}-4 \times$ $111 \times(1413-v) \geqslant 0$, we get
$$
v \geqslant \frac{4 \times 111 \times 1413-750^{2}}{4 \times 111}>0,
$$

which again leads to a contradiction $0>0$, so $t=0$ or 1.
When $t=1$, $b=a+1, c=a+3$, then $111 a^{2}-850 a+1312=0$, this equation has no integer solutions.
When $t=0$, $b=a+1, c=a+2$, then $111 a^{2}-950 a+1211=0$, i.e., $(a-7)(111 a-173)=0$.

Since $a$ is a positive integer, so $a=7, b=8, c=9$, thus $y=1987$.