5. Let $n$ be a positive integer. There are $n$ boxes, each containing some crystal balls (empty boxes are allowed). In each operation, you are allowed to select a box, take out two crystal balls from it, discard one, and place the other in another selected box. For a certain initial distribution of crystal balls, if after a finite number of operations (which could be 0) there are no empty boxes, then this initial distribution is called "solvable." Find all such initial distributions that are not solvable, but become solvable if one more crystal ball is added to any box.
5. The desired distribution is: there are a total of $2 n-2$ crystal balls in all the boxes, and the number of crystal balls in each box is even.
Number the boxes from $1 \sim n$, and use the array
$$
x=\left(x_{1}, x_{2}, \cdots, x_{n}\right)
$$

to represent a distribution state, where $x_{i}(i=1,2, \cdots, n)$ represents the number of crystal balls in the $i$-th box. For the distribution $x$, define
$$
D(x)=\sum_{i=1}^{n}\left[\frac{x_{i}-1}{2}\right] \text {, }
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Equation (1) can be rewritten as
$$
D(x)=\frac{1}{2} N(x)-n+\frac{1}{2} O(x),
$$

where $N(x)$ represents the total number of balls in all the boxes in the distribution $x$, and $O(x)$ represents the number of boxes in the distribution $x$ that contain an odd number of balls.

Notice that each operation either leaves the value of $D(x)$ unchanged (if the operation is performed on a box with an odd number of crystal balls) or decreases the value of $D(x)$ by 1 (if the operation is performed on a box with an even number of crystal balls). For any distribution without empty boxes, $D(x)$ is non-negative, and similarly, for any solvable distribution, $D(x)$ is also non-negative.

On the other hand, for a distribution $x$ where $D(x)$ is non-negative, $D(x) \geqslant 0$ indicates that
$$
\sum_{m_{i}>0} m_{i}\left(m_{i}=\left[\frac{x_{i}-1}{2}\right]\right)
$$

is not less than the number of empty boxes. For each $i$ satisfying $m_{i}>0$, $m_{i}$ steps can be performed to move the balls from the $i$-th box to $m_{i}$ empty boxes, thus ensuring that all boxes contain at least one ball.

Since $N(x)$ and $O(x)$ have the same parity, a distribution $x$ is solvable if and only if $O(x) \geqslant 2 n-N(x)$; and a distribution $x$ is unsolvable if and only if $O(x) \leqslant 2 n-2-N(x)$. In particular, any distribution with a total of $2 n-1$ balls is solvable, and any distribution with a total of $2 n-2$ balls is unsolvable if and only if all the boxes contain an even number of balls.

Let $x^{\prime}$ be the distribution obtained by adding one ball to a box in the distribution $x$. Then
$$
O\left(x^{\prime}\right)=O(x)+1 \text { or } O(x)-1 .
$$

If the distribution $x$ is unsolvable and the distribution $x^{\prime}$ is solvable, then $O(x) \leqslant 2 n-2-N(x)$,
$$
O\left(x^{\prime}\right) \geqslant 2 n-N\left(x^{\prime}\right)=2 n-1-N(x) \text {. }
$$

Thus, $O\left(x^{\prime}\right)=O(x)+1$, meaning the added ball must be placed in a box with an even number of balls. This indicates that only when all the boxes contain an even number of balls, adding a ball to any box will make it solvable, and we have
$$
\begin{array}{l}
0=O(x) \leqslant 2 n-2-N(x), \\
1=O\left(x^{\prime}\right) \geqslant 2 n-1-N(x) .
\end{array}
$$

Therefore, $N(x)=2 n-2$.