5. Given nine positive integers $a_{1}, a_{2}, \cdots, a_{9}$ (allowing repetition) that satisfy: for any $1 \leqslant i<j<k \leqslant 9$, there exists an $l(1 \leqslant l \leqslant 9)$ different from $i, j, k$, such that
$$
a_{i}+a_{j}+a_{k}+a_{l}=100 \text {. }
$$

Find the number of ordered 9-tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ that satisfy the condition.
(Provided by He Yijie)
5. For the positive integer tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$, arrange $a_{1}, a_{2}, \cdots, a_{9}$ in non-decreasing order as
$$
b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{9} \text {. }
$$

From the conditions, we know that there exist $l \in\{4,5, \cdots, 9\}$ and $l^{\prime} \in\{1,2, \cdots, 6\}$, such that
$$
b_{1}+b_{2}+b_{3}+b_{l}=b_{l^{\prime}}+b_{7}+b_{8}+b_{9}=100 \text {. }
$$

Notice that,
$$
b_{l^{\prime}} \geqslant b_{1}, b_{7} \geqslant b_{2}, b_{8} \geqslant b_{3}, b_{9} \geqslant b_{l_{1}} \text {. }
$$

Combining with equation (1), we know that the inequalities in conclusion (2) are all equalities.
Thus, $b_{2}=b_{3}=\cdots=b_{8}$.
Therefore, let $\left(b_{1}, b_{2}, \cdots, b_{8}, b_{9}\right)=(x, y, \cdots, y, z)$, where $x \leqslant y \leqslant z$.

From the conditions, we know that the value of $b_{l}$ that satisfies $x+y+z+b_{l}=100$ can only be $y$, i.e.,
$$
x+2 y+z=100 \text {. }
$$
(1) When $x=y=z=25$, we have
$$
\left(b_{1}, b_{2}, \cdots, b_{9}\right)=(25,25, \cdots, 25) \text {. }
$$

In this case, we get one tuple $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$.
(2) When exactly one of $x, z$ is equal to $y$, let the other be $w$.
From equation (3), we know $w+3 y=100$.
This condition is also sufficient. At this time, $y$ can take
$$
1,2, \cdots, 24,26,27, \cdots, 33
$$

These 32 different values, and each $y$ value corresponds to one tuple $\left(b_{1}, b_{2}, \cdots, b_{9}\right)$, and thus, to nine different tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$, totaling $32 \times 9=288$ tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$.
(3) When $x<y<z$, from the conditions, we know that there exists some $b_{l} \in\{x, y, z\}$, such that $3 y+b_{l}=100$, comparing with equation (3) we get $y+b_{l}=x+z$, then it must be that $b_{l}=y$.
Thus, $y=25, x+z=50$.
This condition is also sufficient. At this time, for $x=1,2$, $\cdots, 24$, each $x$ value corresponds to one tuple $\left(b_{1}, b_{2}, \cdots, b_{9}\right)$, and thus, to $9 \times 8=72$ different tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$, totaling $24 \times 72=1728$ tuples $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$.
In summary, the number of tuples that satisfy the conditions is
$$
1+288+1728=2017 .
$$