> **Answer:** No this is not true.

At least if I am not mistaken, then $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$ is a counter example for $m=5$. In this case we have:
$$\begin{align}
U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0
\end{array}\right) \\
U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0
\end{array}\right) \\
U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1
\end{array}\right) \\
U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2
\end{array}\right) \\
U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\
5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4
\end{array}\right) 
\end{align}$$

Each $U(k)$ has only $14$ entries $\geq k$.