Find all positive integers $n \ge  2$ such that equality  $i+j \equiv C_{n}^{i} +  C_{n}^{j}$ (mod $2$) is true for arbitrary $0 \le  i  \le j  \le n$.
To solve the problem, we need to find all positive integers \( n \ge 2 \) such that the equality \( i + j \equiv \binom{n}{i} + \binom{n}{j} \pmod{2} \) holds for arbitrary \( 0 \le i \le j \le n \).

1. **Claim:**
   There exist such \( n \) if and only if \( \binom{n}{i} \equiv i + 1 \pmod{2} \) for \( 0 < i < n \).

2. **Proof:**
   We start by examining the binomial coefficient modulo 2. Using Lucas' Theorem, we know that:
   \[
   \binom{n}{i} \equiv \prod_{k=0}^{m} \binom{n_k}{i_k} \pmod{2}
   \]
   where \( n_k \) and \( i_k \) are the digits of \( n \) and \( i \) in base 2, respectively.

3. **Simplification:**
   For \( \binom{n}{i} \equiv i + 1 \pmod{2} \), we need to check the parity of \( \binom{n}{i} \). We know that:
   \[
   \binom{n}{i} \equiv i - \binom{n}{0} \equiv i - 1 \equiv i + 1 \pmod{2}
   \]
   This implies that \( \binom{n}{i} \) must be odd if \( i \) is even and even if \( i \) is odd.

4. **Case Analysis:**
   Let \( n = 2^p + q \) where \( 0 < q < 2^p \). Consider the binomial coefficient \( \binom{n}{2^p - 2} \):
   \[
   \binom{n}{2^p - 2} = \frac{(2^p + q) \cdots (q + 3)}{(2^p - 2) \cdots 1}
   \]
   Using Legendre's formula, the \( v_2 \) (2-adic valuation) of the denominator is:
   \[
   v_2(\text{denominator}) = \sum_{i=0}^{\infty} \left\lfloor \frac{2^p - 2}{2^i} \right\rfloor
   \]
   The numerator contains \( 2^p \) numbers, so if \( q + 3 \le 2^p \), then \( \binom{n}{2^p - 2} \) is even. Since \( 2^p - 2 \) is even, we get a contradiction.

5. **Special Cases:**
   - If \( n = 2^p - 1 \), it does not satisfy the condition.
   - If \( n = 2^p - 2 \), we have:
     \[
     \binom{2^p - 2}{i} = \frac{(2^p - 2) \cdots (2^p - i - 1)}{i!}
     \]
     Note that \( \text{ord}_2(2^p - i) = \text{ord}_2(i) \), so the \( v_2 \) of the numerator is greater than that of the denominator. Thus, \( \binom{2^p - 2}{i} \) is even only if \( i \) is odd. Therefore, \( n = 2^p - 2 \) satisfies the condition.

Conclusion:
The only positive integers \( n \ge 2 \) that satisfy the given condition are of the form \( n = 2^p - 2 \).

The final answer is \( \boxed{ n = 2^p - 2 } \).