For which $n$ can it be achieved that among the sums of the form $\pm 1 \pm 2 \pm \ldots \pm n$, the number 100 appears?
I. solution. An integer is even if and only if its opposite is also even; thus, regardless of the sign, the terms, and thus the entire sum, have the same parity as the sum $S_{n}=1+2+\ldots+n$. Therefore, if $S_{n}$ is odd, then 100 cannot be represented in the desired form.

Suppose $S_{n}$ is even and greater than 100; then

$$
S_{n}=100+2d, \quad \text{where} \quad d>1,
$$

so 1 does not appear in the representation of $k$; in this case, $k+1=1+(r+(r+1)+(r+2)+\ldots+n)$.

This proves that 100 can be represented in the desired form if and only if $S_{n} \geq 100$, and $S_{n}$ is even. The former condition requires $n \geq 14$, and the latter that $n$ or $n+1$ is divisible by 4.

$$
\text{Kristóf Bérczi (Szeged, Ságvári E. Gymnasium, 10th grade)}
$$

II. solution. We only deal with the case when $n \geq 14$. Since $S_{14}=105$ is odd, we have $n \geq 15$. Notice that

$$
\begin{gathered}
A_{15}=1+2+\ldots+9-10+11+\ldots+15=100, \quad \text{and} \\
A_{16}=1+2+\ldots+7-8+9-10+11+\ldots+16=100
\end{gathered}
$$

The signs of four consecutive natural numbers can always be chosen so that their sum is 0:

$$
a-(a+1)-(a+2)+(a+3)=0
$$

Starting from the sums $A_{15}$ and $A_{16}$, we can use this identity to represent 100 for every $n$ that is of the form $15+4k$ or $16+4k$. From 15 onwards, these are precisely the numbers for which $S_{n}$ is even.

Tamás Birkner (Budapest, Fazekas M. Gymnasium, 8th grade)