26***. In how many ways can the number $n$ be represented as the sum of three positive integer addends, if representations differing only in the order of the addends are considered the same?
26. Arrange the terms of this sum in ascending order. It is clear that the first (smallest) term can vary from 1 to $\left[\frac{n}{3}\right]$, i.e., there are $\left[\frac{n}{3}\right]$ possibilities for it.

If the first term is $x$, then the second cannot be less than $x$ and greater than $\left[\frac{n-x}{2}\right]$ (since it is not less than the first and not greater than the third). Thus, there are $\left[\frac{n-x}{2}\right]-(x-1)$ possibilities for the second term. The third term is uniquely determined by the first two. Since $x$ can take values $1,2, \ldots,\left[\frac{n}{3}\right]-1,\left[\frac{n}{3}\right]$, the number of ways is

$$
\begin{aligned}
& {\left[\frac{n-1}{2}\right]+\left(\left[\frac{n-2}{2}\right]-1\right)+} \\
& \quad+\left(\left[\frac{n-3}{2}\right]-2\right)+\ldots+\left\{\left[\frac{n-\left[\frac{n}{3}\right]}{2}\right]-\left(\left[\frac{n}{3}\right]-1\right)\right\}
\end{aligned}
$$

Now let's transform this expression. Let the remainder of the division of $n$ by 3 be $k$ ( $k$ can be 0, 1, or 2 ); then $n=3 l+k$, and our expression is equal to the following:

$$
\begin{aligned}
{\left[\frac{3 l+k-1}{2}\right]+} & {\left[\frac{3 l+k-2}{2}\right]+\left[\frac{3 l+k-3}{2}\right]+\ldots } \\
& \ldots+\left[\frac{3 l+k-l}{2}\right]-\{1+2+\ldots+(l-1)\}
\end{aligned}
$$

The expression in curly braces is $\frac{l(l-1)}{2}$. Now let's calculate the sum $S$ of the remaining terms. Rewrite it in reverse order; then it will take the form

$$
\begin{aligned}
S=\left[\frac{2 l+k}{2}\right]+ & {\left[\frac{2 l+k+1}{2}\right]+\ldots } \\
& \ldots+\left[\frac{2 l+k+(l-2)}{2}\right]+\left[\frac{2 l+k+(l-1)}{2}\right]
\end{aligned}
$$

This sum consists of $l$ terms. Using the formula

$$
\left[\frac{m}{2}\right]=\frac{m}{2}-\frac{1}{4}+\frac{(-1)^{m}}{4}
$$

(see the solution to problem 21a)), we can write this sum as:

$$
\begin{gathered}
S=\left(\frac{2 l+k}{2}-\frac{1}{4}+\frac{(-1)^{k}}{4}\right)+\left(\frac{2 l+k}{2}+\frac{1}{2}-\frac{1}{4}+\frac{(-1)^{k+1}}{4}\right)+ \\
+\left(\frac{2 l+k}{2}+\frac{2}{2}-\frac{1}{4}+\frac{(-1)^{k+2}}{4}\right)+\ldots \\
\cdots+\left(\frac{2 l+k}{2}+\frac{l-1}{2}-\frac{1}{4}+\frac{(-1)^{k+l-1}}{4}\right)
\end{gathered}
$$

Rearranging the terms in a different order, we get:

$$
\begin{aligned}
S=\frac{l(2 l+k)}{2}- & \frac{l}{4}+\left(\frac{1}{2}+1+\frac{3}{2}+\cdots+\frac{l-1}{2}\right)+ \\
& +\left(\frac{(-1)^{k}}{4}+\frac{(-1)^{k+1}}{4}+\ldots+\frac{(-1)^{k+l-1}}{4}\right)
\end{aligned}
$$

Denote the sum $\frac{(-1)^{k}}{4}+\frac{(-1)^{k+1}}{4}+\ldots+\frac{(-1)^{k+l-1}}{4}$ by $\varepsilon$. Clearly, $\varepsilon$ can be $+\frac{1}{4}, 0$ or $-\frac{1}{4}$ (see the solution to problem 21a)). Further,

$$
\frac{1}{2}+1+\frac{3}{2}+\ldots+\frac{l-1}{2}=\frac{l(l-1)}{4}
$$

(this is the sum of the terms of an arithmetic progression).

Thus, finally,

$$
\begin{aligned}
& S-\frac{l(l-1)}{2}=\frac{l(2 l+k)}{2}-\frac{l}{4}+\frac{l(l-1)}{4}+\varepsilon-\frac{l(l-1)}{2}= \\
&=\frac{l(4 l+2 k-1-l+1)}{4}+\varepsilon=\frac{l(3 l+2 k)}{4}+\varepsilon \\
& \text { But }
\end{aligned}
$$

$$
\begin{aligned}
\frac{l(3 l+2 k)}{4}+\varepsilon & =\frac{3 l(3 l+2 k)}{12}+\varepsilon=\frac{(3 l+k-k)(3 l+k+k)}{12}+\varepsilon= \\
& =\frac{(n-k)(n+k)}{12}+\varepsilon=\frac{n^{2}-k^{2}}{12}+\varepsilon=\frac{n^{2}}{12}-\frac{k^{2}}{12}+\varepsilon
\end{aligned}
$$

Note that if $k=0$, then $\frac{k^{2}}{12}=0$, and $\varepsilon$ can be 0 or $+\frac{1}{4}$; if $k=1$, then $\frac{k^{2}}{12}=\frac{1}{12}$, and $\varepsilon$ can be 0 or $-\frac{1}{4}$. Therefore, $-\frac{k^{2}}{12}+\varepsilon$ is between $-\frac{1}{3}$ and $\frac{1}{4}$; consequently, the number of ways to represent the number $n$ as the sum of three positive integers is the nearest integer to $\frac{n^{2}}{12}$, i.e., $\left(\frac{n^{2}}{12}\right)$.