Consider the following system of $10$ equations in $10$ real variables $v_1, \ldots, v_{10}$:
\[v_i = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} \qquad (i = 1, \ldots, 10).\]
Find all $10$-tuples $(v_1, v_2, \ldots , v_{10})$ that are solutions of this system.
1. Let \( Q = v_1^2 + v_2^2 + \ldots + v_{10}^2 \). Rewrite the given equations as:
   \[
   v_i = 1 + \frac{6v_i^2}{Q} \quad \text{for} \quad i = 1, \ldots, 10.
   \]
   Rearrange to get:
   \[
   (v_i - 1)Q = 6v_i^2.
   \]
   Since \( Q \) is non-negative, we can assume \( v_i > 1 \) for all \( i \). Dividing both sides by \( v_i - 1 \), we get:
   \[
   Q = \frac{6v_i^2}{v_i - 1}.
   \]

2. Let \( S_i = \frac{6v_i^2}{v_i - 1} \). Since \( Q \) is the same for all \( i \), we have \( S_i = S_j \) for all \( i, j \). This implies:
   \[
   \frac{6v_i^2}{v_i - 1} = \frac{6v_j^2}{v_j - 1}.
   \]
   Simplifying, we get:
   \[
   v_i^2 (v_j - 1) = v_j^2 (v_i - 1).
   \]
   Expanding and rearranging terms, we obtain:
   \[
   v_i^2 v_j - v_i^2 = v_j^2 v_i - v_j^2.
   \]
   \[
   v_i^2 v_j - v_j^2 v_i = v_i^2 - v_j^2.
   \]
   Factoring out common terms, we get:
   \[
   v_i v_j (v_i - v_j) = (v_i - v_j)(v_i + v_j).
   \]
   For \( v_i \neq v_j \), we can divide both sides by \( v_i - v_j \) to get:
   \[
   v_i v_j = v_i + v_j.
   \]
   Rearranging, we find:
   \[
   v_i (v_j - 1) = v_j.
   \]
   Thus:
   \[
   v_i = \frac{v_j}{v_j - 1}.
   \]

3. Since \( S_i = S_j \) for all \( i, j \), there must be some \( v_i = v_j = a \) where \( a \) is a real value greater than 1. For any other \( v_k \), we must have \( v_k = a \) or \( v_k = \frac{a}{a-1} \).

4. Partition the set \( \{v_1, v_2, \ldots, v_{10}\} \) into two subsets: one of size \( k \) consisting of \( a \)'s and the other of size \( 10-k \) consisting of \( \frac{a}{a-1} \)'s. Let \( r = \frac{1}{a-1} \). Then:
   \[
   Q = ka^2 + (10-k) \left( \frac{a}{a-1} \right)^2.
   \]
   Substituting \( Q = \frac{6a^2}{a-1} \), we get:
   \[
   ka^2 + (10-k) \left( \frac{a}{a-1} \right)^2 = \frac{6a^2}{a-1}.
   \]
   Simplifying, we get:
   \[
   (10-k)r^2 - 6r + k = 0.
   \]
   The discriminant of this quadratic equation must be non-negative:
   \[
   36 \geq 4k(10-k).
   \]
   Solving this inequality, we find:
   \[
   k = 0, 1, 9, 10.
   \]

5. Check the cases:
   - If \( k = 0 \), then all \( v_i \) are the same, and we get:
     \[
     v_i = \frac{8}{5} \quad \text{for all} \quad i.
     \]
   - If \( k = 1 \), then \( r = \frac{1}{3} \), so \( a = 4 \) and \( \frac{a}{a-1} = \frac{4}{3} \). Thus:
     \[
     (v_1, v_2, \ldots, v_{10}) = (\frac{4}{3}, \frac{4}{3}, \ldots, \frac{4}{3}, 4) \quad \text{or any permutation}.
     \]

The final answer is \( \boxed{ (\frac{8}{5}, \frac{8}{5}, \ldots, \frac{8}{5}) } \) or \((\frac{4}{3}, \frac{4}{3}, \ldots, \frac{4}{3}, 4)\) and their permutations.