4. The numbers are different natural numbers $a_{1}, a_{2}, \ldots, a_{100}$. For $i=1,2, \ldots, 100$, the number $b_{i}$ is obtained by adding to the number $a_{i}$ the greatest common divisor of the other 99 numbers. How many of the numbers $b_{i}$ can be different at the minimum?
4. It can be 99, for example, for $a_{1}=1$ and $a_{i}=2 i-2$ for $i=2, \ldots, 100$. Then we have $b_{1}=b_{2}=3$ and $b_{i}=a_{i}+1=2 i-1$ for $i \geqslant 3$.

Let's now prove that among the numbers $b_{i}$ there must always be at least 99 different ones. Taking, without loss of generality, $a_{1}i$ and $i, j \neq k$ it holds that $a_{j}-a_{i} \geqslant d_{k} \geqslant d_{i}$, i.e., $b_{j}>a_{j} \geqslant b_{i}$. Therefore, all 99 numbers $b_{i}$ for $i \neq k$ are different.