4. (10 points) The distance between locations $A$ and $B$ in the Smurf Kingdom is 2010 meters. The king sends a messenger from $A$ to $B$ every minute. The speed of the 1st messenger is 1 meter/minute, and the speed of each subsequent messenger is 1 meter/minute faster than the previous one, until the 2010th messenger is sent. Which messengers will arrive at $B$ at the same time?
【Analysis】From the first messenger to the last messenger delivering messages from $A$ to $B$, those who can arrive at $B$ simultaneously must be divisors of 2010. The conclusion can be drawn from this.
【Solution】Let the $m$-th and $n$-th messengers arrive at $B$ simultaneously. Then from $t_{m}=n$ we get $\frac{2010}{V_{\mathbb{m}}}=n$, and then $\frac{2010}{\mathrm{~m}}=n$,

Thus, $m n=2010=2 \times 3 \times 5 \times 67$, which shows that the messengers arriving at $B$ simultaneously appear in pairs,
There are 8 pairs in total: (1, 2010), (2, 1005), (3, 670), (5, 402), (6, 335), (10, 201), (15, 134), (30, 67);

Answer: There are (1,2010), (2,1005), (3,670), (5,402), (6,335), (10,201), (15,134), (30,67) these 8 pairs of messengers arriving simultaneously.