Solution:
Let the $1984$ consecutive positive integers be $n, n+1, n+2, \ldots, n+1983$.

The sum of their squares is:
$$S = n^2 + (n+1)^2 + (n+2)^2 + \cdots + (n+1983)^2$$

This can be written as:
$$S = \sum_{k=0}^{1983} (n+k)^2 = \sum_{k=0}^{1983} (n^2 + 2nk + k^2) = 1984 n^2 + 2n \sum_{k=0}^{1983} k + \sum_{k=0}^{1983} k^2$$

We have:
$$\sum_{k=0}^{m} k = \frac{m(m+1)}{2}$$
$$\sum_{k=0}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$$
For $m = 1983$:
$$\sum_{k=0}^{1983} k = \frac{1983 \times 1984}{2}$$
$$\sum_{k=0}^{1983} k^2 = \frac{1983 \times 1984 \times 3967}{6}$$

So,
$$S = 1984 n^2 + 2n \cdot \frac{1983 \times 1984}{2} + \frac{1983 \times 1984 \times 3967}{6}$$
$$= 1984 n^2 + 1983 \times 1984 n + \frac{1983 \times 1984 \times 3967}{6}$$

Factor $1984$:
$1984 = 2^6 \times 31$.

Let us factor $S$ as follows:
$$S = 1984 (n^2 + 1983 n) + \frac{1983 \times 1984 \times 3967}{6}$$

Let us factor $1983$ and $3967$:
$1983 = 3 \times 661$
$3967 = 2 \times 1983 + 1$

Let us write $S$ as:
$$S = 1984 (n^2 + 1983 n + \frac{1983 \times 3967}{6})$$

But $n^2 + 1983 n + \frac{1983 \times 3967}{6}$ is not necessarily an integer, so let's combine all terms over $6$:

$$S = 1984 \left( n^2 + 1983 n + \frac{1983 \times 3967}{6} \right )$$
$$= 1984 \left( \frac{6 n^2 + 6 \times 1983 n + 1983 \times 3967}{6} \right )$$
$$= \frac{1984}{6} \left( 6 n^2 + 6 \times 1983 n + 1983 \times 3967 \right )$$

Let us compute $6 n^2 + 6 \times 1983 n + 1983 \times 3967$:

$6 n^2 + 6 \times 1983 n = 6 n^2 + 11898 n$

$1983 \times 3967 = 1983 \times (2 \times 1983 + 1) = 1983 \times 2 \times 1983 + 1983 = 2 \times 1983^2 + 1983$

$1983^2 = (2000 - 17)^2 = 4,000,000 - 68,000 + 289 = 3,932,289$

$2 \times 1983^2 = 7,864,578$

$2 \times 1983^2 + 1983 = 7,864,578 + 1983 = 7,866,561$

So the numerator is:
$6 n^2 + 11898 n + 7,866,561$

Therefore,
$$S = \frac{1984}{6} (6 n^2 + 11898 n + 7,866,561)$$

Now, $1984/6 = 992/3$, so $S = \frac{992}{3} (6 n^2 + 11898 n + 7,866,561)$

For $S$ to be an integer, $6 n^2 + 11898 n + 7,866,561$ must be divisible by $3$.

But the main point is to show that $S$ cannot be a perfect square for any integer $n$.

Let us consider the sum of $k$ consecutive squares:

It is known that the sum of $k$ consecutive squares cannot be a perfect square unless $k = 1$ or $k = 24$ (see, for example, Mordell's result).

But $1984$ is not $1$ or $24$.

Alternatively, consider the sum modulo $4$.

The square of any integer is congruent to $0$ or $1$ modulo $4$.

Let us compute $S$ modulo $4$.

Since $1984$ is divisible by $4$, the sum $S$ is divisible by $4$.

But more generally, we can use the following result:

Suppose $k$ is even and $k \equiv 2 \pmod{4}$, then the sum of $k$ consecutive squares cannot be a perfect square.

But $1984 \equiv 0 \pmod{4}$, so this does not help.

Alternatively, consider modulo $3$.

The squares modulo $3$ are $0$ and $1$.

But the key is that for $k > 1$, the sum of $k$ consecutive squares cannot be a perfect square except for $k = 24$ (see Dickson, History of the Theory of Numbers, Vol. II, p. 413).

Therefore, the sum of the squares of $1984$ consecutive positive integers cannot be a perfect square.