Let $m$ be a positive integer. Find the number of real solutions of the equation
$$|\sum_{k=0}^{m} \binom{2m}{2k}x^k|=|x-1|^m$$
Given the equation:
\[ \left| \sum_{k=0}^{m} \binom{2m}{2k} x^k \right| = |x-1|^m \]

We need to find the number of real solutions for this equation.

### Step 1: Analyze the case \( x = 0 \)
If \( x = 0 \), the left-hand side (LHS) becomes:
\[ \left| \sum_{k=0}^{m} \binom{2m}{2k} \cdot 0^k \right| = \left| \binom{2m}{0} \cdot 0^0 \right| = 1 \]
The right-hand side (RHS) becomes:
\[ |0-1|^m = 1^m = 1 \]
Thus, \( x = 0 \) is a solution.

### Step 2: Consider \( x > 0 \)
Let \( x = y^2 \) where \( y > 0 \). The equation becomes:
\[ \left| \sum_{k=0}^{m} \binom{2m}{2k} y^{2k} \right| = |y^2 - 1|^m \]

Using the binomial theorem, we know:
\[ (1+y)^{2m} + (1-y)^{2m} = 2 \sum_{k=0}^{m} \binom{2m}{2k} y^{2k} \]
Thus, the equation simplifies to:
\[ \frac{(1+y)^{2m} + (1-y)^{2m}}{2} = 2 |1+y|^m |1-y|^m \]

This implies:
\[ ((1+y)^m \pm (1-y)^m)^2 = 0 \]
The only solution to this is \( y = 0 \), which implies \( x = 0 \). Since \( x = 0 \) is not a positive solution, there are no positive solutions.

### Step 3: Consider \( x < 0 \)
Let \( x = -y^2 \) where \( y > 0 \). The equation becomes:
\[ \left| \sum_{k=0}^{m} \binom{2m}{2k} (-y^2)^k \right| = |(-y^2) - 1|^m \]

Using the binomial theorem, we know:
\[ (1+iy)^{2m} + (1-iy)^{2m} = 2 \sum_{k=0}^{m} \binom{2m}{2k} (-1)^k y^{2k} \]
Thus, the equation simplifies to:
\[ \left| \frac{(1+iy)^{2m} + (1-iy)^{2m}}{2} \right| = 2 |1+iy|^m |1-iy|^m \]

This implies:
\[ ((1+iy)^m \pm (1-iy)^m)^2 = 0 \]
Thus:
\[ (1+iy)^{2m} = (1-iy)^{2m} \]

After some algebraic manipulation, we find:
\[ y = \tan \frac{k\pi}{2m} \]
for \( k \in \{1, 2, \ldots, m-1\} \).

Thus, the solutions are:
\[ x = -\tan^2 \frac{k\pi}{2m} \]
for \( k \in \{1, 2, \ldots, m-1\} \).

### Conclusion
There are exactly \( m-1 \) solutions for \( x < 0 \).

The final answer is \( \boxed{ m-1 } \) solutions: \( x = -\tan^2 \frac{k\pi}{2m} \) for \( k \in \{1, 2, \ldots, m-1\} \).