Let $ n$ be a composite natural number and $ p$ a proper divisor of $ n.$ Find the binary representation of the smallest natural number $ N$ such that 

\[ \frac{(1 \plus{} 2^p \plus{} 2^{n\minus{}p})N \minus{} 1}{2^n}\]

is an integer.
To solve the problem, we need to find the smallest natural number \( N \) such that 

\[
\frac{(1 + 2^p + 2^{n-p})N - 1}{2^n}
\]

is an integer. This means that \((1 + 2^p + 2^{n-p})N - 1\) must be divisible by \(2^n\). In other words, we need to find \( N \) such that:

\[
(1 + 2^p + 2^{n-p})N \equiv 1 \pmod{2^n}
\]

This implies that \( N \) is the multiplicative inverse of \( 1 + 2^p + 2^{n-p} \) modulo \( 2^n \).

### Case 1: \( \frac{n}{p} \) is odd

If \( \frac{n}{p} \) is odd, we can find the inverse as follows:

\[
N = 1 - 2^p + 2^{2p} - 2^{3p} + \cdots - 2^{n-2p} + 2^n
\]

To verify, we compute:

\[
(1 + 2^p + 2^{n-p})(1 - 2^p + 2^{2p} - 2^{3p} + \cdots - 2^{n-2p} + 2^n)
\]

Expanding the product:

\[
(1 + 2^p + 2^{n-p})(1 - 2^p + 2^{2p} - 2^{3p} + \cdots - 2^{n-2p})
\]

\[
= (1 + 2^p)(1 - 2^p + 2^{2p} - 2^{3p} + \cdots - 2^{n-2p}) + 2^{n-p}
\]

\[
= 1 - 2^p + 2^{2p} - 2^{3p} + \cdots - 2^{n-2p} + 2^p - 2^{2p} + 2^{3p} - \cdots + 2^{n-p} + 2^{n-p}
\]

\[
= 1 + 2^{n-p} - 2^{n-p} = 1 \pmod{2^n}
\]

Thus, the inverse is correct.

### Case 2: \( \frac{n}{p} \) is even

If \( \frac{n}{p} \) is even, the inverse is:

\[
N = 1 - 2^p + 2^{2p} - 2^{3p} + \cdots + 2^{n-2p} - 2^{n-p+1} + 2^n
\]

To verify, we compute:

\[
(1 + 2^p + 2^{n-p})(1 - 2^p + 2^{2p} - 2^{3p} + \cdots + 2^{n-2p} - 2^{n-p+1} + 2^n)
\]

Expanding the product:

\[
(1 + 2^p + 2^{n-p})(1 - 2^p + 2^{2p} - 2^{3p} + \cdots + 2^{n-2p} - 2^{n-p+1})
\]

\[
= (1 + 2^p)(1 - 2^p + 2^{2p} - 2^{3p} + \cdots + 2^{n-2p} - 2^{n-p+1}) + 2^{n-p}
\]

\[
= 1 - 2^p + 2^{2p} - 2^{3p} + \cdots + 2^{n-2p} - 2^{n-p+1} + 2^p - 2^{2p} + 2^{3p} - \cdots - 2^{n-p+1} + 2^{n-p}
\]

\[
= 1 + 2^{n-p} - 2^{n-p+1} + 2^{n-p} = 1 + 2^{n-p+1} - 2^{n-p+1} = 1 \pmod{2^n}
\]

Thus, the inverse is correct.

### Conclusion

The smallest natural number \( N \) such that 

\[
\frac{(1 + 2^p + 2^{n-p})N - 1}{2^n}
\]

is an integer is given by the binary representation of the inverse found in the respective cases.

The final answer is the binary representation of the inverse \( \boxed{ N } \) as described in the respective cases.