Consecutive positive odd numbers form an arithmetic sequence with a difference of 2. Let's denote the smallest number in the sequence by $a$, and the number of terms by $n$. It is not entirely clear from the problem statement whether the case $n=1$ is allowed. In what follows, we will assume that $n>1$. (At the grading, neither the exclusion nor the allowance of the $n=1$ case resulted in any point deduction.)

Using the sum formula for an arithmetic sequence, we have

$$
1995=\frac{a+(a+2 \cdot(n-1))}{2} \cdot n=(a+n-1) \cdot n
$$

Thus, the number of decompositions is the number of ways to factorize 1995 as a product of integers in the above manner. Therefore,

$$
n|1995, \quad a+n-1| 1995
$$

and since $a \geq 1$, we have

$$
a+n-1 \geq n
$$

which implies

$$
n \leq \sqrt{1995}
$$

Factoring 1995 into prime factors, we get:

$$
1995=3 \cdot 5 \cdot 7 \cdot 19
$$

Thus, the positive divisors are: $1,3,5,7,19,3 \cdot 5,3 \cdot 7,3 \cdot 19,5 \cdot 7,5 \cdot 19,7 \cdot 19,3 \cdot 5 \cdot 7,3 \cdot 5 \cdot 19,3 \cdot 7 \cdot 19$, $5 \cdot 7 \cdot 19,3 \cdot 5 \cdot 7 \cdot 19$. Among these, those not greater than $\sqrt{1995}$ are: $1,3,5,7,19,3 \cdot 5,3 \cdot 7,5 \cdot 7$.

We have already excluded the case $n=1$, so there are 7 possibilities for $n$, corresponding to the following decompositions:

$$
n=3
$$

$n=5$

$n=7$

$n=15$

$n=19$

$n=21$

$n=35$

$$
\begin{aligned}
663+665+667 & =1995 \\
395+397+\cdots+403 & =1995 \\
279+281+\cdots+291 & =1995 \\
119+121+\cdots+147 & =1995 \\
87+89+\cdots+123 & =1995 \\
75+77+\cdots+115 & =1995 \\
23+25+\cdots+91 & =1995
\end{aligned}
$$

Based on the work of Tibor Zámbó (Kaposvár, Római Katolikus Gimn., II.o.t.)

Note. Many forgot that we are looking for decompositions into positive terms. They did not use the condition $n \leq \sqrt{1995}$, and thus obtained many more decompositions, most of which contained negative terms.
In how many ways can 1995 be decomposed into the sum of consecutive positive odd numbers?