Problem 1473

1. Define the problem and notation:
Let $d$ be a natural number. Given two natural numbers $M$ and $N$ with $d$ digits, $M$ is a friend of $N$ if and only if the $d$ numbers obtained by substituting each one of the digits of $M$ by the digit of $N$ which is in the same position are all multiples of $7$. We need to find all values of $d$ for which the following condition is valid: For any two numbers $M$ and $N$ with $d$ digits, $M$ is a friend of $N$ if and only if $N$ is a friend of $M$.

2. **Express $M$ and $N$ in terms of their digits:**
Let $M = \sum_{k=0}^{d-1} m_k 10^k$ and $N = \sum_{k=0}^{d-1} n_k 10^k$.

3. **Condition for $M$ to be a friend of $N$:**
The statement that $M$ is a friend of $N$ is equivalent to:
$$M + 10^k (n_k - m_k) \equiv 0 \pmod{7} \quad \text{for all } 0 \leq k \leq d-1.$$

4. **Case 1: $d \not\equiv 2 \pmod{7}$:**
- Assume $M$ is a friend of $N$ and vice versa.
- Summing all the equalities, we get:
$$dM + N - M \equiv 0 \pmod{7} \quad \text{and} \quad dN + M - N \equiv 0 \pmod{7}.$$
- Subtracting these two congruences:
$$(d-2)M \equiv (d-2)N \pmod{7} \implies M \equiv N \pmod{7}.$$
- Since $M$ and $N$ are congruent modulo 7, we can write:
$$M + 10^k (n_k - m_k) \equiv 0 \equiv N + 10^k (m_k - n_k) \equiv M + 10^k (m_k - n_k) \pmod{7}.$$
- This implies $M \equiv N \equiv 0 \pmod{7}$.
- However, there exist numbers $M$ and $N$ such that $M$ is a friend of $N$ and $7 \nmid M$. For example, take $M = 10^{d-1}$, $n_{d-1} = 7$, and $n_k \equiv -10^{d-k} \pmod{7}$. If $k < d-1$, then:
$$M + 10^k (n_k - m_k) = 10^d + 10^k n_k \equiv 0 \pmod{7},$$
and if $k = d-1$, then:
$$M + 10^k (n_k - m_k) = 7 \times 10^{d-1} \equiv 0 \pmod{7}.$$
- This shows that $N$ is not necessarily a friend of $M$, contradicting the condition.

5. **Case 2: $d \equiv 2 \pmod{7}$:**
- Assume $M$ is a friend of $N$.
- Summing the equalities $M + 10^k (n_k - m_k) \equiv 0 \pmod{7}$, we get:
$$dM + N - M \equiv 0 \pmod{7} \implies 2M + N - M \equiv M + N \pmod{7} \quad \text{(since \( d \equiv 2 \pmod{7} \))}.$$
- This implies:
$$N + 10^k (m_k - n_k) \equiv -M - 10^k (n_k - m_k) \equiv 0 \pmod{7},$$
so $N$ is a friend of $M$, as desired.

6. Conclusion:
The only values of $d$ for which the condition is valid are those such that $d \equiv 2 \pmod{7}$.

The final answer is $\boxed{ d \equiv 2 \pmod{7} }$.