Can the signs in the expression $\pm 1 \pm 2 \pm \ldots \pm 100$ be chosen such that the value of the expression is exactly 1982?
We will show that in the expression $\pm 1 \pm 2 \pm \ldots \pm 100$, by choosing the signs appropriately, every even number between -5050 and +5050 can be obtained. If all signs are negative, the sum is $-\frac{100 \cdot 101}{2}=-5050$, and if all are positive, the sum is 5050.

We will prove our statement by induction. We already know that -5050 can be obtained this way. We will show that if $u<5050$ is among the sums, then $u+2$ is also among them. Consider the first negative sign from the left in the expression for $u$ (since $u<5050$, such a sign exists). If this sign is in front of 1, changing it to positive increases the sum by two. If it is in front of a number greater than 1, changing this sign to positive and the sign in front of the number one less to negative, the sum again increases by two, thus we have obtained $(u+2)$.

Of course, a number can be represented in multiple different ways. One possible representation of 1982 is:

$$
-1-2-\ldots-5+6-7-8 \ldots-54-55+56+\ldots+100
$$

Jenő Töröcsik (Budapest, Fazekas M. Gyak. Gimn. IV. o. t.)