Solution:

The total number of chips for an $n \times n$ board is equal to
$$\begin{aligned} n \times 0 + 2 \times (n-1) \times 1 + 2 \times (n-2) \times 2 + \cdots + 2 \times 1 \times (n-1) &= \sum_{i=1}^{n} 2 \times (n-i) \times i \\ &= 2\left(n \sum_{i=1}^{n} i - \sum_{i=1}^{n} i^{2}\right) \\ &= 2\left(n \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}\right) \\ &= \frac{1}{3} n^{3} - \frac{1}{3} n \end{aligned}$$
Find $n$ such that $\frac{1}{3} n^{3} - \frac{1}{3} n = 2660$. This leads to $n^{3} - n = 7980$. Note that $n^{3} - n - 7980 = (n-20)(n^{2} + 20n + 399) = 0$ and the only integer satisfying it is $20$.