1101*. Find all natural $n$ for which in the expression

$$
* 1 * 2 * 3 * \ldots * n=n+1
$$

it is possible to replace the asterisks with plus and minus signs so that the resulting equation is true.
$\triangle$ First, let's replace all the stars with plus signs. We get:

$$
1+2+3+\ldots+n=\frac{n(n+1)}{2}
$$

When replacing some of the plus signs with minus signs, the parity of the sum does not change. Therefore, the number $\frac{n(n+1)}{2}$ must have the same parity as the number $n+1$. But then the difference between these numbers is even. Let's find it:

$$
\frac{n(n+1)}{2}-(n+1)=\frac{(n+1)(n-2)}{2}
$$

In this case, the product $(n+1)(n-2)$ will be divisible by 4. Since the factors $n+1$ and $n-2$ have different parities, only one of them is divisible by 4.

So far, we have established the following: if the original equation holds, then $n+1$ or $n-2$ is divisible by 4.

Let's prove the converse. We will consider two possibilities.

1) Suppose $n+1$ is divisible by 4:

$$
n+1=4 k, \quad n=4 k-1 \quad(k \in \mathbb{N})
$$

Transform $n$ as follows:

$$
n=(4 k-4)+(4-1)=4(k-1)+3=4 m+3
$$

where $m=k-1$ and $m$ takes non-negative integer values.

The case $m=0$ is obvious: $-1+2+3=4$. Let $m \geqslant 1$. Arrange the plus and minus signs as follows:

$$
(-1+2+3-4)+(-5+6+7-8)+\ldots+(-(4 m-3)+(4 m-2)+(4 m-1)-4 m)-
$$

$$
-(4 m+1)+(4 m+2)+(4 m+3)=0 \cdot m+4 \cdot m+4=n+1
$$

2) Suppose $n-2$ is divisible by 4:

$$
n-2=4 k, \quad n=4 k+2, \quad(k=0,1,2, \ldots)
$$

240

The case $k=0$ is obvious. Let $k$ be a natural number. Arrange the plus and minus signs in the original expression as follows:

$$
\begin{gathered}
(1+2-3-4)+(5+6-7-8)+\ldots+((4 k-3)+(4 k-2)-(4 k-1)-4 k)+ \\
+(4 k+1)+(4 k+2)=-4 k+(4 k+1)+(4 k+2)=4 k+3=n+1
\end{gathered}
$$

Answer: $n=4 k+2$ or $n=4 k+3(k=0,1,2, \ldots)$