44 *. The sequence of integers $x_{0}, x_{1}, x_{2}, \ldots$ is such that $x_{0}=0$, and $\left|x_{n}\right|=\left|x_{n-1}+1\right|$ for each natural $n$. What is the smallest possible value of the expression $\mid \mathrm{x}_{1}+x_{2}+\ldots$ $\ldots+x_{1975} \mid$?
75.44. Answer. 20. Since $x_{1}^{2}=1, x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \ldots$ $\ldots, x_{1976}^{2}=x_{1975}^{2}+2 x_{1975}+1$, then, adding all these equalities, we get

$$
\sum_{i=1}^{1975} x_{i}=\left(x_{1976}^{2}-1976\right) / 2
$$

Since the closest square to 1976 of an even number ( $x_{1976}$ is even!) is $44^{2}=1936$, then $\left|\sum_{i=1}^{1975} x_{i}\right| \geqslant 20$. The reader can easily verify that this is the exact value, as constructing the sequence $\left\{x_{i}\right\}$ for which $x_{1976}=44$ is quite straightforward.