Solution:

The minimum is achieved by staying on the perimeter of the grid, and the maximum by staying as close to the main diagonal as possible. To see why this is true, tilt the grid 45 degrees:

Now every path must include exactly one number from each row. The $n^{\text{th}}$ row consists of the numbers $k(n+1-k)$ for $k=1,2, \ldots, n$. As a quadratic function of $k$, this expression is greatest when $k=(n+1)/2$, and gets smaller as $k$ gets farther from $(n+1)/2$ in either direction. Thus the smallest number in row $n$ of our tilted grid is the number on either end of the row, which is $n$, and the largest number in row $n$ is the number closest to the center, which is $\left\lfloor\frac{n+1}{2}\right\rfloor \cdot\left\lceil\frac{n+1}{2}\right\rceil$. By staying on the perimeter of the grid, we may select the smallest entry from every row, and so accumulate the smallest possible total. By staying in the middle, we may select the largest entry from every row and accumulate the largest possible total.

All that remains is to calculate these totals in terms of $n$.

The minimum total is
$$\begin{aligned} (1+2+3+\cdots+(n-1))+(n+2n+3n+\cdots+n^{2}) &= (1+2+\cdots+(n-1))+n(1+2+\cdots+n) \\ &= \frac{n(n-1)}{2} + n \cdot \frac{n(n+1)}{2} \\ &= \frac{n\left(n^{2}+2n-1\right)}{2}. \end{aligned}$$

The maximum total is
$$\begin{aligned} 1^{2} + 1 \cdot 2 + 2^{2} + 2 \cdot 3 + \cdots + (n-1) \cdot n + n^{2} &= \left(1^{2} + 2^{2} + \cdots + n^{2}\right) + (1 \cdot 2 + 2 \cdot 3 + \cdots + (n-1) \cdot n) \\ &= \frac{n(n+1)(2n+1)}{6} + 2 \cdot \left[\binom{2}{2} + \binom{3}{2} + \cdots + \binom{n}{2}\right] \\ &= \frac{n(n+1)(2n+1)}{6} + 2\binom{n+1}{3} \\ &= \frac{n(n+1)(2n+1)}{6} + \frac{2(n-1)n(n+1)}{6} \\ &= \frac{n(n+1)(4n-1)}{6} \end{aligned}$$
and we are finished.