4. Given positive integers $a$ and $b$ differ by 120, their least common multiple is 105 times their greatest common divisor. The larger of $a$ and $b$ is $\qquad$
4.225.

Let $(a, b)=d$, and $a=m d, b=n d$, where $m>n, m$ and $n$ are coprime. Thus, the least common multiple of $a$ and $b$ is $m n d$. According to the problem, we have
$$
\left\{\begin{array}{l}
m d-n d=120, \\
\frac{m n d}{d}=105,
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}(m-n) d=2^{3} \times 3 \times 5, \\ m n=3 \times 5 \times 7 .\end{array}\right.
$$
Since $m>n$, from equation (2) we get
$$
\left\{\begin{array}{l}
m=105, \\
n=1;
\end{array} \quad \left\{\begin{array}{l}
m=35, \\
n=3;
\end{array} \quad \left\{\begin{array}{l}
m=21, \\
n=5;
\end{array} \quad \left\{\begin{array}{l}
m=15, \\
n=7 .
\end{array}\right.\right.\right.\right.
$$

According to equation (1), only $\left\{\begin{array}{l}m=15, \\ n=7 .\end{array}\right.$ is valid, and we can find $d=15$.
Therefore, the larger number is $m d=225$.