We alternately color (as a chessboard) the cells of an $n$-type tiles triangle, as below:

We have $b_n = 1 + 2 + \ldots + n = \frac{n(n+1)}{2}$ black cells and $a_n = 1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}$ white cells. In any sequence of cells with the required propriety, their color alternates, hence its length cannot exceed $2(a_n + 1)$. Therefore, we infer that the maximal length of such a sequence is $L \leq n^2 - n + 1$.

To complete the solution, we now provide an example of a sequence in an $n$-type tiles triangle which satisfies the required condition and has length $L = n^2 - n + 1$.
We consider a sequence of cells which begins with the top cell. At each step, we choose the cell below the last cell chosen in the previous step, together with a maximum length sequence of cells situated on the same line, up to one of its ends. For instance, for a 3-type tiles triangle the following succession is a solution:

The length of a sequence as we described before is: $1 + 2 + 4 + \ldots + 2(n - 1) = 1 + n(n - 1) = n^2 - n + 1$.

Therefore, the answer is $L = n^2 - n + 1$.