7. Let $a_{1}, a_{2}, \cdots, a_{10}$ be a permutation of $2000, 2001, \cdots$, 2009, and let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$. Then the probability that the permutation $a_{1}, a_{2}, \cdots, a_{10}$ satisfies “ $S_{i}(1 \leqslant i \leqslant 10)$ are not multiples of 3" is $\qquad$.
7. $\frac{1}{50}$.

Let a permutation of $2000, 2001, \cdots, 2009$ be a basic event $M$. Then the total number of basic events is
$$
N=\mathrm{A}_{10}^{10} \text {. }
$$

Next, we calculate the number of basic events contained in the event $M$.
(1) The first term cannot be a multiple of 3, and the other terms can be multiples of 3. Thus, there are $\mathrm{A}_{9}^{3}$ ways to arrange the multiples of 3;
(2) After removing the multiples of 3, consider the positions of numbers that are congruent to 2 and 1 modulo 3 (replace $a_{i}$ with its remainder when divided by 3):

When $a_{1}=1$, $a_{2}=1, a_{3}=2, a_{4}=1, \cdots \cdots$ In this case, the number of terms containing 1 is more than the number of terms containing 2, which contradicts the given condition;

When $a_{1}=2$, $a_{2}=2, a_{3}=1, \cdots \cdots$ In this case, the condition is satisfied.

In summary, the positions of numbers that are congruent to 2 and 1 modulo 3 are uniquely determined, and their respective arrangements are $\mathrm{A}_{4}^{4}$ and $\mathrm{A}_{3}^{3}$.
Therefore, the number of basic events contained in event $M$ is
$$
m=\mathrm{A}_{9}^{3} \mathrm{~A}_{4}^{4} \mathrm{~A}_{3}^{3} \text {. }
$$

Thus, the required probability $P=\frac{m}{N}=\frac{1}{50}$.