Example 11 Given the equation $\sum_{k=1}^{1950} x_{k}=2000$. Find:
(1) the number of solutions in natural numbers;
(2) the number of solutions in non-negative integers.
Solution: (1) Construct a line segment $AB$, and divide it into 2000 equal parts.

Let $A$ and $B$ be the 0th and 2000th positions, respectively. The other points are numbered from left to right as 1st, 2nd, $\cdots, 1999$th. It is stipulated that if $x_{1}$ is placed at the 2nd position, then $x_{1}=2$, and if $x_{2}$ is placed at the 5th position, then $x_{2}=5-2=3$. Also, place $x_{1950}$ at point $B$. Thus, the problem is equivalent to selecting 1949 positions from the 1st to 1999th positions to place $x_{k}(1 \leqslant k \leqslant 1949)$, so the number of natural number solutions to the equation is $\mathrm{C}_{1999}^{1949}$.
(2) Let $y_{k}=x_{k}+1$, then the problem is equivalent to finding the number of natural number solutions to the equation $\sum_{k=1}^{1951} y_{k}=2000+1950=3950$. According to (1), it is easy to know that the number of non-negative integer solutions to the equation is $\mathrm{C}_{3949}^{1949}$.

Note. Problems involving the number of non-negative integer (or natural number) solutions to multiple linear indeterminate equations are often solved using permutation and combination knowledge.