5. Uncle Qiel Yuemuer every night selects 9 or 10 warriors from 33 warriors to be on duty according to his own wishes. At least how many days are needed to ensure that each warrior has the same number of duty shifts?
The minimum number of days is 7 days, during which each warrior is on duty 2 times.

Let the required minimum number of days be $N$, and each warrior is on duty $n$ times within these $N$ days. We number the warriors sequentially as $1,2, \cdots, 33$, and introduce the number $a_{1 \times}(i=1,2, \cdots, N ; k=1,2, \cdots, 33)$ to represent the following: if the $k$-th warrior is not on duty on the $i$-th day, then $a_{1 x}=0$; if the $k$-th warrior is on duty on the $i$-th day, then $a_{i k}=1$. Let $P_{k}$ be the number of people on duty on the $i$-th day, and $Q_{k}$ be the number of days the $k$-th warrior is on duty. Thus, according to the problem conditions, the following relationships hold:
$$
\begin{array}{l}
9 \leqslant a_{i 1}+a_{\mathrm{i} 2}+\cdots+a_{\mathrm{i} 33}=P_{\mathrm{i}} \leqslant \\
10, i=1,2, \cdots, N \text {; } \\
\text { (1) } \\
a_{1_{k}}+a_{2 k}+\cdots+a_{N k}=Q_{k}=n \text {, } \\
i=1,2, \cdots, \dot{N} \text {; } \\
\text { (2) } \\
\end{array}
$$
\begin{tabular}{|c|c|c|}
\hline Day & On-duty Warrior Numbers & \begin{tabular}{l} 
On-duty
\end{tabular} \\
\hline 1 & $2,3,4,5,6,7,8,9,28,29$ & 10 \\
\hline 2 & $1,2,3,4,5,6,7,8,9$ & 9 \\
\hline 3 & $11,12,13,14,15,16,17,18,30,31$ & 20 \\
\hline 4 & $10,11,12,13,14,15,16,17,18$ & $\frac{9}{10}$ \\
\hline 5 & $20,21,22,23,24,25,26,27,32,33$ & . 0 \\
\hline 6 & $19,20,21,22,23,24,25,26,27$ & $\theta$ \\
\hline 7 & $1,10,19,28,29,30,31,32,33$ & 9 \\
\hline
\end{tabular}

Figure 12
Summing the inequality (1) for $i=1,2, \cdots, N$, we get the inequality $9 N \leqslant S \leqslant 10 N$, where:
$$
\begin{array}{l}
S=\left(a_{11}+a_{12}+\cdots+a_{1,33}\right)+\left(a_{21}\right. \\
\left.+a_{22}+\cdots+a_{2,33}\right)+\cdots+\left(a_{\mathrm{N} 1}+a_{\mathrm{N} 2}+\cdots\right. \\
\left.+a_{\mathrm{N}, 33}\right) .
\end{array}
$$

Summing the equation (2) for $k=1,2, \cdots, 33$, we get the equation $S=33 n$. This implies $N \leqslant 33 n \leqslant 10 N$, or $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. An integer $n$ between $\frac{3}{11} N$ and $\frac{10}{33} N$ does not exist for $N=1,2,3,4,5,6$, but when $N=7$, there is an integer $n=2$ that satisfies the inequality $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. Therefore, the number of duty shifts $n$ is at least 2, and the minimum number of days $N$ is at least 7. The duty schedule (similar to the table in Figure 12) meets the problem's conditions.