[Solution] Clearly, all terms of the sequence are positive integers.
Since when $n$ is even,
$$a_{n+1}=\frac{1}{2} a_{n}<a_{n}$$

Therefore, the smallest term of the sequence must be odd.
Let the smallest term of the sequence be $a_{p}$, then $a_{p}$ is odd, and
$$\begin{array}{l} a_{p+1}=a_{p}+7 \\ a_{p+2}=\frac{1}{2} a_{p+1}=\frac{1}{2}\left(a_{p}+7\right) \end{array}$$

By the minimality of $a_{p}$, we have
$$a_{p} \leqslant a_{p+2}$$

That is,
$$\begin{array}{l} a_{p} \leqslant \frac{1}{2}\left(a_{p}+7\right) \\ a_{p} \leqslant 7 \end{array}$$

Noting that $a_{p}$ is odd. Therefore, $a_{p}$ has only 4 possible values:
$$1,3,5,7$$

Since
$$1994 \equiv-1(\bmod 3)$$

Therefore
$$1994^{1995} \equiv(-1)^{1995}=-1 \equiv 2(\bmod 3)$$

That is, there exists a positive integer $t$, such that
$$1994^{1995}=3 t+2$$

Since 1994 is even, $t$ is even, let $t=2 s$, we get
$$1994^{1995}=6 s+2$$

Here $s$ is a positive integer.
Thus, we have
$$\begin{aligned} a_{1} & =1993^{1994^{1995}}=1993^{6 s+2} \\ & =\left(1993^{6}\right)^{s} \cdot 1993^{2} \\ & \equiv 1^{s} \cdot(-2)^{2}(\bmod 7) \\ & \equiv 4(\bmod 7) \end{aligned}$$

From $a_{1} \equiv 4(\bmod 7)$, we know that the remainders of the terms of this sequence when divided by 7 can only be $4,2,1$. Therefore, $a_{p}$ cannot be $3,5,7$, the value of the smallest term of the sequence
$$a_{p}=1$$