Example 2 Find four consecutive integers, each of which is divisible by $2^{2}, 3^{2}, 5^{2}$, and $7^{2}$ respectively.
Let the four consecutive integers be $x-1, x, x+1, x+2$. According to the requirements, they should satisfy:
$$\begin{array}{ll}
x-1 \equiv 0\left(\bmod 2^{2}\right), & x \equiv 0\left(\bmod 3^{2}\right) \\
x+1 \equiv 0\left(\bmod 5^{2}\right), & x+2 \equiv 0\left(\bmod 7^{2}\right)
\end{array}$$

Thus, this is a problem of solving a system of congruences, where
$$m_{1}=2^{2}, \quad m_{2}=3^{2}, \quad m_{3}=5^{2}, \quad m_{4}=7^{2}$$

These are pairwise coprime, satisfying the conditions of Theorem 1. We have $M_{1}=3^{2} 5^{2} 7^{2}, M_{2}=2^{2} 5^{2} 7^{2}, M_{3}=$ $2^{2} 3^{2} 7^{2}, M_{4}=2^{2} 3^{2} 5^{2}$. From $M_{1} \equiv 1 \cdot 1 \cdot 1 \equiv 1\left(\bmod 2^{2}\right)$, we know
$$1=M_{1} M_{1}^{-1} \equiv M_{1}^{-1}\left(\bmod 2^{2}\right)$$

Thus, we can take $M_{1}^{-1}=1$. From $M_{2} \equiv 10^{2} \cdot 7^{2} \equiv 1 \cdot 4 \equiv 4\left(\bmod 3^{2}\right)$, we know
$$1 \equiv M_{2} M_{2}^{-1} \equiv 4 M_{2}^{-1}\left(\bmod 3^{2}\right)$$

Thus, we can take $M_{2}^{-1}=-2$. From $M_{3} \equiv 2^{2} \cdot 21^{2} \equiv 2^{2} \cdot 4^{2} \equiv-11\left(\bmod 5^{2}\right)$, we know
$$\begin{array}{l}
1 \equiv M_{3} M_{3}^{-1} \equiv-11 M_{3}^{-1}\left(\bmod 5^{2}\right) \\
2 \equiv-22 M_{3}^{-1} \equiv 3 M_{3}^{-1}\left(\bmod 5^{2}\right) \\
16 \equiv 24 M_{3}^{-1} \equiv-M_{3}^{-1}\left(\bmod 5^{2}\right)
\end{array}$$

Thus, we can take $M_{3}^{-1}=9$. From $M_{4}=(-13)(-24) \equiv 3 \cdot 6 \equiv 18\left(\bmod 7^{2}\right)$, we know
$$\begin{array}{l}
1 \equiv M_{4} M_{4}^{-1} \equiv 18 M_{4}^{-1}\left(\bmod 7^{2}\right) \\
3 \equiv 54 M_{4}^{-1} \equiv 5 M_{4}^{-1}\left(\bmod 7^{2}\right) \\
30 \equiv 50 M_{4}^{-1} \equiv M_{4}^{-1}\left(\bmod 7^{2}\right)
\end{array}$$

Thus, we can take $M_{4}^{-1}=-19$. Therefore, by Theorem 1, we have
$$\begin{aligned}
x \equiv & 3^{2} \cdot 5^{2} \cdot 7^{2} \cdot 1 \cdot 1+2^{2} \cdot 5^{2} \cdot 7^{2} \cdot(-2) \cdot 0 \\
& +2^{2} \cdot 3^{2} \cdot 7^{2} \cdot 9 \cdot(-1)+2^{2} \cdot 3^{2} \cdot 5^{2} \\
& \cdot(-19) \cdot(-2)\left(\bmod 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2}\right) \\
x \equiv & 11025-15876+34200 \equiv 29349(\bmod 44100)
\end{aligned}$$

Therefore, there are infinitely many sets of four consecutive integers that satisfy the requirements, which are
$$\begin{array}{c}
29348+44100 t, \quad 29349+44100 t \\
29350+44100 t, \quad 29351+44100 t \\
t=0, \pm 1, \pm 2, \cdots
\end{array}$$

The smallest such set of four consecutive positive integers is
$$29348,29349,29350,29351 .$$