The first term of the arithmetic progression is given as 61, so \(a = 61\).

The second term is 57, so the common difference \(d\) is \(57 - 61 = -4\).

The formula for the sum of the first \(n\) terms of an arithmetic progression is:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Given that the sum \(S_n = n\), we have:

\(n = \frac{n}{2} [2 \times 61 + (n-1)(-4)]\)

Simplifying inside the brackets:

\(n = \frac{n}{2} [122 + (n-1)(-4)]\)

\(n = \frac{n}{2} [122 - 4n + 4]\)

\(n = \frac{n}{2} [126 - 4n]\)

Multiply both sides by 2 to eliminate the fraction:

\(2n = n(126 - 4n)\)

\(2n = 126n - 4n^2\)

Rearrange to form a quadratic equation:

\(4n^2 - 124n = 0\)

Factor out \(n\):

\(n(4n - 124) = 0\)

So, \(n = 0\) or \(4n - 124 = 0\).

Since \(n\) must be positive, solve \(4n - 124 = 0\):

\(4n = 124\)

\(n = 31\)

Thus, the value of the positive integer \(n\) is 31.