144 From the $n \times n$ array
$$
\left[\begin{array}{ccccc}
1 & 2 & 3 & \cdots & n \\
n+1 & n+2 & n+3 & \cdots & 2 n \\
2 n+1 & 2 n+2 & 2 n+3 & \cdots & 3 n \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
(n-1) n+1 & (n-1) n+2 & (n-1) n+3 & \cdots & n^{2}
\end{array}\right]
$$

select $x_{1}$, delete the row and column containing $x_{1}$, then select a number $x_{2}$ from the new array, and delete the row and column containing $x_{2}$. Continue this process until only one number $x_{n}$ remains. Then the value of $x_{1}+x_{2}+\cdots+x_{n}$ is . $\qquad$
$144 \frac{n\left(n^{2}+1\right)}{2}$. Let $x_{i},(1 \leqslant i \leqslant n)$, belong to the $r_{i}$ row and $s_{i}$ column of the array, then
$$
x_{i}=\left(r_{i}-1\right) n+s_{i} ;(1 \leqslant i \leqslant n) .
$$

Thus,
$$
\sum_{i=1}^{n} x_{i}=n \sum_{i=1}^{n} r_{i}-n^{2}+\sum_{i=1}^{n} s_{i} .
$$

Since $\left\{r_{1}, \cdots, r_{n}\right\}$ and $\left\{s_{1}, \cdots, s_{n}\right\}$ are both permutations of $\{1,2, \cdots, n\}$, we have
$$
\sum_{i=1}^{n} r_{i}=\sum_{i=1}^{n} s_{i}=\sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$

Therefore,
$$
\sum_{i=1}^{n} x_{i}=\frac{n^{2}(n+1)}{2}-n^{2}+\frac{n(n+1)}{2}=\frac{n\left(n^{2}+1\right)}{2} .
$$