The Mathematician's Agony

Call the combination of one forward move and one backward move a 'walk'. It is clear that after the $k$ th walk, assuming he hasn't fallen into the pit yet, the mathematician will be $k$ meters from the starting point.

Consider what happens during the $n$ th walk. The mathematician starts at position $n - 1$ metres (that's where the previous walk ended), walks $n$ metres forward to reach position $2n - 1$ metres, then walks back $n - 1$ metres to position $n$ metres. So he'll fall into the pit the first time $2n - 1 > 2018$ , i.e when $n = 1010$ .

By the time he completes the $1009$ th walk, the mathematician's walked $\ 1 + 2 + 3 + .... + 1009 \$ metres forward and $\ 0 + 1 + 2 + .... + 1008 \$ metres backward for a total distance of

$\dfrac{(1009)(1010)}{2} + \dfrac{(1008)(1009)}{2} = 1018081 \$ metres.

He starts the $1010$ th walk at position $1009$ metres, and would have walked forward $1010$ metres but falls into the pit after walking $1009$ metres.

Thus the distance he walks before falling into the pit is $\ 1018081 + 1009 = \boxed{1019090 m}$