Can you explain the problem more clearly? What defines if two friends are sitting next to each other? Isn't the first condition given, because obviously there are more seats than people per bus?

Sorry, I made a typo. They're not sitting next to each other. There are as many seats as people, they are all taken.

Let $n$ and $m$ the number of people in each bus and $N=m+n$ the total number of people. The probability that two friends are seated in the first bus is $\frac{ \binom {n} {2}}{ \binom {m+n} {2}}$ and the probability that they are in the second is $\frac{ \binom {m} {2}}{ \binom {m+n} {2}}$ so we have : $$\frac{\binom {m} {2}+\binom {m} {2}}{\binom {m+n} {2}}=\frac{1}{2}$$ a) The last relation is equivalent to $N=m+n=(m-n)^2$ thus $N$ is a perfect square. b) The equation $m+n=(m-n)^2$ is a quadratic and has a solution $n=m+\frac{1-\sqrt{8m+1}}{2}$ thus $8m+1$ is an odd perfect square. Let $8m+1=(2a+1)^2=4a^2+4a+1$ thus $m=\frac{a(a+1)}{2}$ is a triangular number $n=\frac{a(a-1)}{2}$ is a triangular number too.